2-6 Nonlinear Inequalities
Solve each inequality.
1. (x + 4)(x − 2) ≤ 0
SOLUTION: Let f (x) = (x + 4)(x − 2). f (x) has real zeros at x = −4 and x = 2. Set up a sign chart. Substitute an x-value in each
test interval into the polynomial to determine if f (x) is positive or negative at that point.
The solutions of (x + 4)(x − 2) ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can
see that the solution set is [–4, 2].
2. (x − 6)(x + 1) > 0
SOLUTION: Let f (x) = (x − 6)(x + 1). f (x) has real zeros at x = 6 and x = −1. Set up a sign chart. Substitute an x-value in each
test interval into the polynomial to determine if f (x) is positive or negative at that point.
The solutions of (x − 6)(x + 1) > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the
solution set is (– , –1) (6, ).
3. (3x + 1)(x – 8) ≥ 0
SOLUTION: Let f (x) = (3x + 1)(x – 8). f (x) has real zeros at x = 8 and
. Set up a sign chart. Substitute an x-value in each
test interval into the polynomial to determine if f (x) is positive or negative at that point.
The solutions of (3x + 1)(x – 8) ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can
see that the solution set is
.
4. (x – 4)(–2x + 5) < 0
SOLUTION: Let f (x) = (x – 4)(–2x + 5). f (x) has real zeros at x = 4 and x =
. Set up a sign chart. Substitute an x-value in each
test interval into the polynomial to determine if f (x) is positive or negative at that point.
The solutions of (x – 4)(–2x + 5) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that
the solution set is
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5. (4 – 6y)(2y + 1) < 0
SOLUTION: .
Page 1
The solutions of (3x + 1)(x – 8) ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can
2-6 see
Nonlinear
Inequalities
that the solution
set is
.
4. (x – 4)(–2x + 5) < 0
SOLUTION: Let f (x) = (x – 4)(–2x + 5). f (x) has real zeros at x = 4 and x =
. Set up a sign chart. Substitute an x-value in each
test interval into the polynomial to determine if f (x) is positive or negative at that point.
The solutions of (x – 4)(–2x + 5) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that
the solution set is
.
5. (4 – 6y)(2y + 1) < 0
SOLUTION: Let f (y) = (4 – 6y)(2y + 1). f (y) has real zeros at y =
and . Set up a sign chart. Substitute n y-value in
each test interval into the polynomial to determine if f (y) is positive or negative at that point.
The solutions of (4 – 6y)(2y + 1) < 0 are y-values such that f (y) is negative. From the sign chart, you can see that
the solution set is
.
6. 2x3 – 9x2 – 20x + 12 ≤ 0
SOLUTION: 3
2
Let f (x) = 2x – 9x – 20x + 12. By using synthetic division, it can be determined that x = −2 is a rational zero.
2
The remaining quadratic factor (2x − 13x + 6) can be written as (2x − 1)(x − 6). A factored form of f (x) is f (x) =
(x + 2)(2x − 1)(x − 6). f (x) has real zeros at x = −2, x =
, and x = 6. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
3
2
The solutions of 2x – 9x – 20x + 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart,
you can see that the solution set is
.
7. –8x3 – 30x2 – 18x < 0
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SOLUTION: 3
2
Let f (x) = –8x – 30x – 18x. A factored form of f (x) is f (x) = −2x(4x + 3)(x + 3). f (x) has real zeros at x = 0,
Page 2
3
2
The solutions of 2x – 9x – 20x + 12 ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart,
2-6 you
Nonlinear
Inequalities
can see that
the solution set is
.
7. –8x3 – 30x2 – 18x < 0
SOLUTION: 3
2
Let f (x) = –8x – 30x – 18x. A factored form of f (x) is f (x) = −2x(4x + 3)(x + 3). f (x) has real zeros at x = 0,
, and x = −3. Set up a sign chart. Substitute an x-value in each test interval into the factored polynomial to
determine if f (x) is positive or negative at that point.
3
2
The solutions of –8x – 30x – 18x < 0 are x-values such that f (x) is negative. From the sign chart, you can see that
the solution set is
.
8. 5x3 – 43x2 + 72x + 36 > 0
SOLUTION: 3
2
Let f (x) = 5x – 43x + 72x + 36. By using synthetic division, it can be determined that x = 3 is a rational zero.
2
The remaining quadratic factor (5x − 28x − 12) can be written as (5x + 2)(x − 6). A factored form of f (x) is f (x) =
(x − 3)(5x + 2)(x − 6). f (x) has real zeros at x = 3,
, and x = 6. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
3
2
The solutions of 5x – 43x + 72x + 36 > 0 are x-values such that f (x) is positive. From the sign chart, you can see
that the solution set is
.
9. x2 + 6x > –10
SOLUTION: 2
2
2
Write x + 6x > –10 as x + 6x + 10 > 0 and let f (x) = x + 6x + 10. Use a graphing calculator to graph f (x).
2
f(x) has no real zeros, so there are no sign changes. The solutions of x + 6x + 10 > 0 are x-values such that f (x) is
2
positive. Because f (x) is positive for all real values of x, the solution set of x + 6x > –10 is all real numbers or (–
eSolutions Manual - Powered by Cognero
Page 3
, ).
10. 2x2 ≤ –x – 4
3
2
The solutions of 5x – 43x + 72x + 36 > 0 are x-values such that f (x) is positive. From the sign chart, you can see
2-6 that
Nonlinear
the solutionInequalities
set is
.
9. x2 + 6x > –10
SOLUTION: 2
2
2
Write x + 6x > –10 as x + 6x + 10 > 0 and let f (x) = x + 6x + 10. Use a graphing calculator to graph f (x).
2
f(x) has no real zeros, so there are no sign changes. The solutions of x + 6x + 10 > 0 are x-values such that f (x) is
2
positive. Because f (x) is positive for all real values of x, the solution set of x + 6x > –10 is all real numbers or (–
, ).
10. 2x2 ≤ –x – 4
SOLUTION: 2
2
2
Write 2x ≤ −x − 4 as 2x + x + 4 ≤ 0 and let f (x) = 2x + x + 4. Use a graphing calculator to graph f (x).
2
f(x) has no real zeros, so there are no sign changes. The solutions of 2x + x + 4 ≤ 0 are x-values such that f (x) is
2
negative or equal to 0. Because f (x) is positive for all real values of x, the solution set of 2x ≤ –x – 4 is
.
11. 4x2 + 8 ≤ 5 – 2x
SOLUTION: 2
2
2
Write 4x + 8 ≤ 5 − 2x as 4x + 2x + 3 ≤ 0 and let f (x) = 4x + 2x + 3. Use a graphing calculator to graph f (x).
2
f(x) has no real zeros, so there are no sign changes. The solutions of 4x + 2x + 3 ≤ 0 are x-values such that f (x) is
2
negative or equal to 0. Because f (x) is positive for all real values of x, the solution set of 4x + 8 ≤ 5 – 2x is
.
12. 2x2 + 8x ≥ 4x – 8
SOLUTION: 2
eSolutions
Manual
- Powered
Write
2x +
8x ≥ 4xby–Cognero
8 as 2x
2
2
2
+ 4x + 8 ≥ 0 and let f (x) = 2x + 4x + 8. A factored form of f (x) is f (x) = 2(x + 2)Page
.f 4
(x) has a real zero at x = −2. Set up a sign chart. Substitute an x-value in each test interval into the factored
polynomial to determine if f (x) is positive or negative at that point.
2
f(x) has no real zeros, so there are no sign changes. The solutions of 4x + 2x + 3 ≤ 0 are x-values such that f (x) is
2-6 Nonlinear Inequalities
2
negative or equal to 0. Because f (x) is positive for all real values of x, the solution set of 4x + 8 ≤ 5 – 2x is
.
12. 2x2 + 8x ≥ 4x – 8
SOLUTION: 2
2
2
2
Write 2x + 8x ≥ 4x – 8 as 2x + 4x + 8 ≥ 0 and let f (x) = 2x + 4x + 8. A factored form of f (x) is f (x) = 2(x + 2) . f
(x) has a real zero at x = −2. Set up a sign chart. Substitute an x-value in each test interval into the factored
polynomial to determine if f (x) is positive or negative at that point.
2
The solutions of 2x + 4x + 8 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can
see that the solution set is (– , ).
13. 2b 2 + 16 ≤ b 2 + 8b
SOLUTION: 2
2
2
2
2
Write 2b + 16 ≤ b + 8b as b − 8b + 16 ≤ 0 and let f (b) = b − 8b + 16. A factored form of f (b) is f (b) = (b − 4) .
f(b) has a real zero at b = 4. Set up a sign chart. Substitute a b-value in each test interval into the factored
polynomial to determine if f (b) is positive or negative at that point.
2
The solutions of b − 8b + 16 ≤ 0 are b-values such that f (b) is negative or equal to 0. From the sign chart, you can
see that the solution set is {4}.
14. c2 + 12 ≤ 3 – 6c
SOLUTION: 2
2
2
2
Write c + 12 ≤ 3 – 6c as c + 6c + 9 ≤ 0 and let f (c) = c + 6c + 9. A factored form of f (c) is f (c) = (c + 3) . f (c)
has a real zero at c = −3. Set up a sign chart. Substitute a c-value in each test interval into the factored polynomial to
determine if f (c) is positive or negative at that point.
2
The solutions of c + 6c + 9 ≤ 0 are c-values such that f (c) is negative or equal to 0. From the sign chart, you can
see that the solution set is {−3}.
15. –a 2 ≥ 4a + 4
SOLUTION: 2
2
2
2
Write –a ≥ 4a + 4 as −a − 4a − 4 ≥ 0 and let f (a) = −a − 4a − 4. A factored form of f (a) is f (a) = −(a + 2) . f
(a) has a real zero at a = −2. Set up a sign chart. Substitute an a-value in each test interval into the factored
polynomial to determine if f (a) is positive or negative at that point.
2
The solutions of −a − 4a − 4 ≥ 0 are a-values such that f (a) is positive or equal to 0. From the sign chart, you can
see that the solution set is {−2}.
16. 3d 2 + 16 ≥ −d 2 + 16d
eSolutions
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SOLUTION: 2
2
Page 5
2
2
Write 3d + 16 ≥ −d + 16d as 4d − 16d + 16 ≥ 0 and let f (d) = 4d − 16d + 16. A factored form of f (d) is f (d) = 4
2
2
solutions ofInequalities
−a − 4a − 4 ≥ 0 are a-values such that f (a) is positive or equal to 0. From the sign chart, you can
2-6 The
Nonlinear
see that the solution set is {−2}.
16. 3d 2 + 16 ≥ −d 2 + 16d
SOLUTION: 2
2
2
2
Write 3d + 16 ≥ −d + 16d as 4d − 16d + 16 ≥ 0 and let f (d) = 4d − 16d + 16. A factored form of f (d) is f (d) = 4
2
(d − 2) . f (d) has a real zero at d = 2. Set up a sign chart. Substitute an d-value in each test interval into the
factored polynomial to determine if f (d) is positive or negative at that point.
2
The solutions of 4d − 16d + 16 ≥ 0 are d-values such that f (d) is positive or equal to 0. From the sign chart, you can
see that the solution set is (– , ).
17. BUSINESS A new company projects that its first-year revenue will be r(x) = 120x – 0.0004x2 and the start-up
cost will be c(x) = 40x + 1,000,000, where x is the number of products sold. The net profit p that they will make the
first year is equal to p = r − c. Write and solve an inequality to determine how many products the company must sell
to make a profit of at least $2,000,000.
SOLUTION: 2
Substitute r(x) = 120x – 0.0004x and c(x) = 40x + 1,000,000 into p = r − c.
Substitute p = 2,000,000 to write an inequality to determine how many products the company must sell to make a
profit of at least $2,000,000.
2
2,000,000 ≤ −0.0004x + 80x − 1,000,000
2
2
Solve the inequality for x. Write the inequality as −0.0004x + 80x − 3,000,000 ≥ 0. Let f (x) = −0.0004x + 80x −
3,000,000 and use a graphing calculator to graph f (x). Use the CALC menu to locate the zeros of f (x).
2
f(x) has zeros at x = 50,000 and x = 150,000. The solutions of −0.0004x + 80x − 3,000,000 ≥ 0 are x-values such
that f (x) is positive or equal to 0. From the graph, you can see that the solution set is [50,000, 150,000] or 50,000 ≤ x
≤ 150,000. So, the company must sell 50,000 ≤ x ≤ 150,000 units to make a profit of at least $2,000,000.
Solve each inequality.
18. >3
SOLUTION: eSolutions Manual - Powered by Cognero
Page 6
2-6
2
f(x) has zeros at x = 50,000 and x = 150,000. The solutions of −0.0004x + 80x − 3,000,000 ≥ 0 are x-values such
that
f (x) is positive or equal to 0. From the graph, you can see that the solution set is [50,000, 150,000] or 50,000 ≤ x
Nonlinear
Inequalities
≤ 150,000. So, the company must sell 50,000 ≤ x ≤ 150,000 units to make a profit of at least $2,000,000.
Solve each inequality.
18. >3
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator,
, and
denominator, x = −4. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
positive or negative.
are x-values such that f (x) is positive. From the sign chart, you can see that the
The solutions of
solution set is
19. .
≤ 1
SOLUTION: Let f (x) =
. There are no zeros because the numerator has no real zeros. The undefined point of the inequality
eSolutions
Manual
Powered
by Cognero
is the
zero -of
the denominator,
x = 5. Create a sign chart. Then choose and test x-values in each interval to
determine if f (x) is positive or negative.
Page 7
are x-values such that f (x) is positive. From the sign chart, you can see that the
The solutions of
solution set is
.
2-6 Nonlinear Inequalities
19. ≤ 1
SOLUTION: Let f (x) =
. There are no zeros because the numerator has no real zeros. The undefined point of the inequality
is the zero of the denominator, x = 5. Create a sign chart. Then choose and test x-values in each interval to
determine if f (x) is positive or negative.
≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that
the solution set is (–
, 5). x = 5 is not part of the solution set because the original inequality is undefined at x = 5.
20. The solutions of
≥ 4
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator, x =
, and
denominator, x = 6. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
positive or negative.
eSolutions Manual - Powered by Cognero
Page 8
The solutions of
≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you can see that
the solution set is (–
, 5). x = 5 is not part of the solution set because the original inequality is undefined at x = 5.
2-6 Nonlinear Inequalities
20. ≥ 4
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator, x =
, and
denominator, x = 6. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
positive or negative.
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
The solutions of
that the solution set is
. x = 6 is not part of the solution set because the original inequality is undefined at x =
6.
21. <6
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator,
denominator,
x = –3.
eSolutions
Manual - Powered
by Create
Cognero a
positive or negative.
sign chart. Then choose and test x-values in each interval to determine if f (x) is
, and
Page 9
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
The solutions of
that the solution set is
. x = 6 is not part of the solution set because the original inequality is undefined at x =
2-6 Nonlinear Inequalities
6.
21. <6
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator,
, and
denominator, x = –3. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
positive or negative.
are x-values such that f (x) is negative. From the sign chart, you can see that the
The solutions of
solution set is
22. .
<5
SOLUTION: Let
and denominator,
. The zeros and undefined points of the inequality are the zeros of the numerator,
. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
eSolutions Manual - Powered by Cognero
positive or negative.
,
Page 10
are x-values such that f (x) is negative. From the sign chart, you can see that the
The solutions of
2-6 solution
Nonlinear
set is Inequalities
22. .
<5
SOLUTION: Let
. The zeros and undefined points of the inequality are the zeros of the numerator,
and denominator,
,
. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
positive or negative.
are x-values such that f (x) is negative. From the sign chart, you can see that the
The solutions of
solution set is
23. .
≥ –3
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator, x =
, and
eSolutions Manual - Powered by Cognero
denominator, x =
positive or negative.
. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
Page 11
are x-values such that f (x) is negative. From the sign chart, you can see that the
The solutions of
2-6 solution
Nonlinear
set is Inequalities
23. .
≥ –3
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator, x =
, and
denominator, x =
. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is
positive or negative.
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
The solutions of
that the solution set is
undefined at x =
24. .x=
is not part of the solution set because the original inequality is .
≤ 6
SOLUTION: eSolutions Manual - Powered by Cognero
Page 12
that the solution set is
.x=
is not part of the solution set because the original inequality is 2-6 undefined
Nonlinear
at x Inequalities
= .
24. ≤ 6
SOLUTION: Let f (x) =
and
. The zeros and undefined points of the inequality are the zeros of the numerator, x = 4
, and denominator, x = 3 and x = −1. Create a sign chart. Then choose and test x-values in each interval
to determine if f (x) is positive or negative.
The solutions of
can see that the solution set is
≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you
. x = −1 and x = 3 are not part of the solution set
because the original inequality is undefined at x = −1 and x = 3.
25. ≤ 4
SOLUTION: eSolutions Manual - Powered by Cognero
Page 13
≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you
The solutions of
. x = −1 and x = 3 are not part of the solution set
can see that the solution set is
2-6 Nonlinear Inequalities
because the original inequality is undefined at x = −1 and x = 3.
25. ≤ 4
SOLUTION: Let
. The zeros and undefined points of the inequality are the zeros of the numerator, x =
,
and denominator, x = −3 and x = 1. Create a sign chart. Then choose and test x-values in each interval to determine
if f (x) is positive or negative.
The solutions of
are x-values such that f (x) is negative or equal to 0. From the sign chart, you can
see that the solution set is
. x = −3 and x = 1 are not part of the solution set because the original
inequality is undefined at x = −3 and x = 1.
26. SOLUTION: eSolutions Manual - Powered by Cognero
Page 14
The solutions of
are x-values such that f (x) is negative or equal to 0. From the sign chart, you can
. x = −3 and x = 1 are not part of the solution set because the original
see that the solution set is
2-6 Nonlinear Inequalities
inequality is undefined at x = −3 and x = 1.
26. SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator,
and x = −5, and denominator, x = −4. Create a sign chart. Then choose and test x-values in each interval to
determine if f (x) is positive or negative.
The solutions of
can see that the solution set is
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you
. x = −4 is not part of the solution set because the original
inequality is undefined at x = −4.
27. SOLUTION: eSolutions Manual - Powered by Cognero
Page 15
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you
The solutions of
. x = −4 is not part of the solution set because the original
can see that the solution set is
2-6 Nonlinear Inequalities
inequality is undefined at x = −4.
27. SOLUTION: Let f (x) =
and x =
. The zeros and undefined points of the inequality are the zeros of the numerator, x = 4
, and denominator, x = 3. Create a sign chart. Then choose and test x-values in each interval to
determine if f (x) is positive or negative.
The solutions of
the solution set is
< 0 are x-values such that f (x) is negative. From the sign chart, you can see that
.
28. CHARITY The Key Club at a high school is having a dinner as a fundraiser for charity. A dining hall that can
accommodate 80 people will cost $1000 to rent. If each ticket costs $20 in advance or $22 the day of the dinner, and
the same number of people bought tickets in advance as bought the day of the dinner, write and solve an inequality
to determine the minimum number of people that must attend for the club to make a profit of at least $500.
SOLUTION: Let x be the total number of people that bought a ticket. Since the same number of people bought tickets in advance
as bought the day of the dinner,
people bought in advance and bought the day of the dinner. To find the profit, multiply the ticket prices by the amount of people that bought each type of ticket and subtract the initial cost to rent
the hall.
eSolutions Manual - Powered by Cognero
Page 16
Write an inequality to determine the minimum number of people that must attend for the club to make a profit of at
least $500.
< 0 are x-values such that f (x) is negative. From the sign chart, you can see that
The solutions of
2-6 the
Nonlinear
solution setInequalities
is
.
28. CHARITY The Key Club at a high school is having a dinner as a fundraiser for charity. A dining hall that can
accommodate 80 people will cost $1000 to rent. If each ticket costs $20 in advance or $22 the day of the dinner, and
the same number of people bought tickets in advance as bought the day of the dinner, write and solve an inequality
to determine the minimum number of people that must attend for the club to make a profit of at least $500.
SOLUTION: Let x be the total number of people that bought a ticket. Since the same number of people bought tickets in advance
as bought the day of the dinner,
people bought in advance and bought the day of the dinner. To find the profit, multiply the ticket prices by the amount of people that bought each type of ticket and subtract the initial cost to rent
the hall.
Write an inequality to determine the minimum number of people that must attend for the club to make a profit of at
least $500.
Solve for x.
Thus, 72 people must attend the dinner for the club to make a profit of at least $500.
29. PROM A group of friends decides to share a limo for prom. The cost of rental is $750 plus a $25 fee for each occupant. There is a minimum of two passengers, and the limo can hold up to 14 people. Write and solve an
inequality to determine how many people can share the limo for less than $120 per person.
SOLUTION: Let x be the total number of people that share the limo. The total cost of the limo is $750 + $25x. The cost per
person is the total cost divided by the total number of people or
. Write an inequality to determine how
many people can share the limo for less than $120 per person.
< 120
Solve for x.
When more than 8 people share the limo, the cost per person will be less than $120 per person. Thus, 8 to 14 people
are needed to be able to share the limo for less than $120 per person.
Find the domain of each expression.
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eSolutions
Page 17
SOLUTION: 2
more than
8 people share the limo, the cost per person will be less than $120 per person. Thus, 8 to 14 people
2-6 When
Nonlinear
Inequalities
are needed to be able to share the limo for less than $120 per person.
Find the domain of each expression.
30. SOLUTION: 2
For the expression to be defined, x + 5x + 6 ≥ 0.
2
Let f (x) = x + 5x + 6. A factored form of f (x) is f (x) = (x + 3)(x + 2). f (x) has real zeros at x = −3 and x = −2. Set
up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive
or negative at that point.
2
The solutions of x + 5x + 6 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
that the solution set is (– , –3] [−2, ).
31. SOLUTION: 2
For the expression to be defined, x − 3x − 40 ≥ 0.
2
Let f (x) = x − 3x −40. A factored form of f (x) is f (x) = (x − 8)(x + 5). f (x) has real zeros at x = 8 and x = −5. Set
up a sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive
or negative at that point.
2
The solutions of x − 3x −40 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can
see that the solution set is (– , –5] [8, ).
32. SOLUTION: 2
For the expression to be defined, 16 − x ≥ 0.
2
Let f (x) = 16 − x . A factored form of f (x) is f (x) = (4 − x)(4 + x). f (x) has real zeros at x = 4 and x = −4. Set up a
sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
2
The solutions of 16 − x ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
that the solution set is [−4, 4].
33. SOLUTION: 2
For the expression to be defined, x − 9 ≥ 0.
2
Let f (x) = x − 9. A factored form of f (x) is f (x) = (x − 3)(x + 3). f (x) has real zeros at x = 3 and x = −3. Set up a
sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
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2
Page 18
The solutions of x − 9 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that
2
solutions ofInequalities
16 − x ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
2-6 The
Nonlinear
that the solution set is [−4, 4].
33. SOLUTION: 2
For the expression to be defined, x − 9 ≥ 0.
2
Let f (x) = x − 9. A factored form of f (x) is f (x) = (x − 3)(x + 3). f (x) has real zeros at x = 3 and x = −3. Set up a
sign chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
2
The solutions of x − 9 ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see that
the solution set is (– , –3] [3, ).
34. SOLUTION: For the expression to be defined,
Let f (x) =
≥ 0.
. A factored form of f (x) is f (x) =
. The zeros and undefined points of the
inequality are the zeros of the numerator, x = 0, and denominator, x = −5 and x = 5. Set up a sign chart. Substitute an
x-value in each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
The solutions of
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
that the solution set is (−5, 0] (5, ). x = −5 and x = 5 are not part of the solution set because the original
inequality is undefined at x = −5 and x = 5.
35. SOLUTION: 2
2
For the expression to be defined, the denominator of the radicand, 36 − x , cannot equal 0. Let f (x) = 36 − x . Set
this function equal to 0 and solve for x.
2
36 − x = 0 at x = 6 and x = −6. Thus, the domain of the expression is (–
, –6)
(−6, 6)
(6, ).
Find the solution set of f (x) – g(x) ≥ 0.
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36. Page 19
2-6 Nonlinear
Inequalities
2
36 − x = 0 at x = 6 and x = −6. Thus, the domain of the expression is (–
, –6)
(−6, 6)
(6, ).
Find the solution set of f (x) – g(x) ≥ 0.
36. SOLUTION: f(x) − g(x) ≥ 0 for values of x that result in values for f (x) that are greater than or equal to the values of g(x). This
occurs when x is (–∞, –1] or [2, ∞). Thus, the solution set is (–∞, –1] ∪ [2, ∞).
37. SOLUTION: f(x) − g(x) ≥ 0 for values of x that result in values for f (x) that are greater than or equal to the values of g(x). This
occurs when x is [–4, 3]. Thus, the solution set is [−4, 3].
38. SALES A vendor sells hot dogs at each school sporting event. The cost of each hot dog is $0.38 and the cost of each bun is $0.12. The vendor rents the hot dog cart that he uses for $1000. If he wants his costs to be less than his
profits after selling 400 hot dogs, what should the vendor charge for each hot dog?
SOLUTION: The vendor wants his costs c to be less than his profit p after selling 400 hot dogs, or p > c. The cost of each hot
dog is $0.38 + $0.12 or $0.50. The total cost of selling 400 hot dogs is 400 · $0.50, or $200, plus the cost of renting the hot dog cart. The profit from selling 400 hot dogs is the product of 400 and the selling price x of a hot dog. Set up
an inequality that represents this situation and solve for x.
The hot dog vendor must charge more than $3 in order for his costs to be less than his profits after selling 400 hot
dogs.
39. PARKS AND RECREATION A rectangular playing field for a community park is to have a perimeter of 112 feet and an area of at least 588 square feet.
a. Write an inequality that could be used to find the possible lengths to which the field can be constructed.
b. Solve the inequality you wrote in part a and interpret the solution.
c. How does the inequality and solution change if the area of the field is to be no more than 588 square feet?
Interpret this new solution in the context of the situation.
SOLUTION: a. The formula for the perimeter of a rectangle is P = 2
eSolutions Manual - Powered by Cognero
+ 2w. Substitute P = 112 and solve for w.
Page 20
hot dog vendor
must charge more than $3 in order for his costs to be less than his profits after selling 400 hot
2-6 The
Nonlinear
Inequalities
dogs.
39. PARKS AND RECREATION A rectangular playing field for a community park is to have a perimeter of 112 feet and an area of at least 588 square feet.
a. Write an inequality that could be used to find the possible lengths to which the field can be constructed.
b. Solve the inequality you wrote in part a and interpret the solution.
c. How does the inequality and solution change if the area of the field is to be no more than 588 square feet?
Interpret this new solution in the context of the situation.
SOLUTION: a. The formula for the perimeter of a rectangle is P = 2
+ 2w. Substitute P = 112 and solve for w.
The formula for the area of a rectangle is lw = A. Substitute w = 56 − and A = 588 to write an inequality that
could be used to find the possible lengths to which the field can be constructed.
2
Thus, an inequality that could be used is (56 – ) ≥ 588 or –
+ 56 – 588 ≥ 0.
2
2
b. Solve –
+ 56 – 588 ≥ 0. Let f ( ) = –
+ 56 – 588. A factored form of f ( ) is f ( ) = (− + 14)( −
42). f ( ) has real zeros at = 14 and = 42. Set up a sign chart. Substitute a -value in each test interval into
the factored polynomial to determine if f ( ) is positive or negative at that point.
2
The solutions of –
+ 56 – 588 ≥ 0 are l-values such that f (l) is positive or equal to 0. From the sign chart, you
can see that the solution set is [14, 42]. The length of the playing field is at least 14 and at most 42 feet.
c. If the area of the field is to be no more than 588 square feet, the inequality becomes 0 < (56 – ) ≤ 588. Notice that the area of the field must be greater than 0.
Start by solving 0 < (56 – ). Since the length, width, and area of the rectangle must be positive, 0 < < 56.
2
The solutions of (56 – ) ≤ 588 or –
+ 56 – 588 ≤ 0 are -values such that f ( ) is negative or equal to 0.
From the sign chart, you can see that the solution set is (– , 14] [42, ).
Since 0 < < 56, the solution set is (0, 14] [42, 56). The area of the playing field must be greater than 0 square feet but at most 588 square feet, so 0 < ≤ 14 or 42 ≤ < 56.
Solve each inequality.
−
> 3
40. SOLUTION: For the inequality to be defined, 9y + 19 ≥ 0 and 6y − 5 ≥ 0. Solve both inequalities for y.
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The domain of y must be restricted to y ≥ Page 21
. Solve the original inequality.
2
2-6
The solutions of (56 – ) ≤ 588 or –
+ 56 – 588 ≤ 0 are -values such that f ( ) is negative or equal to 0.
From the sign chart, you can see that the solution set is (– , 14] [42, ).
Since
0 < < 56, the solution set is (0, 14] [42, 56). The area of the playing field must be greater than 0 square Nonlinear
Inequalities
feet but at most 588 square feet, so 0 < ≤ 14 or 42 ≤ < 56.
Solve each inequality.
40. −
> 3
SOLUTION: For the inequality to be defined, 9y + 19 ≥ 0 and 6y − 5 ≥ 0. Solve both inequalities for y.
The domain of y must be restricted to y ≥ . Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (y) = 9(y − 5)(y − 9). f (y) has real zeros at y = 5 and y = 9. Set up a sign chart. Substitute a y-value in each test
interval into the factored polynomial to determine if f (y) is positive or negative at that point.
The solutions of 9(y − 5)(y − 9) > 0 are y-values such that f (y) is positive. From the sign chart, you can see that the
solution set is (–∞, 5) ∪ (9, ∞). Recall that y ≥ . Create a chart that includes this restriction. When solving
inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using
the original inequality. Substitute a y-value in each test interval into the original inequality to determine if f (y) is a
solution.
Thus, the solution set is
eSolutions
41. Manual - Powered by Cognero
−
SOLUTION: ≤ 4
.
Page 22
Thus, the solution set is
.
2-6 Nonlinear Inequalities
41. −
≤ 4
SOLUTION: For the inequality to be defined, 4x + 4 ≥ 0 and x − 4 ≥ 0. Solve both inequalities for x.
The domain of x must be restricted to x ≥4. Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (x) = (9x − 40)(x − 8). f (x) has real zeros at x =
and x = 8. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
The solutions of (9x − 40)(x − 8) ≤ 0 are x-values such that f (x) is negative or equal to 0. From the sign chart, you
can see that the solution set is
. Recall that x ≥ 4. Create a chart that includes this restriction. When solving inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using
the original inequality. Substitute an x-value in each test interval into the original inequality to determine if f (x) is a
solution.
Thus, the solution set is
42. −
.
≥ 7
SOLUTION: eSolutions
Manual - Powered by Cognero
For the inequality to be defined, 12y + 72 ≥ 0 and 6y − 11 ≥ 0. Solve both inequalities for y.
Page 23
2-6 Thus,
Nonlinear
Inequalities
the solution
set is
42. −
.
≥ 7
SOLUTION: For the inequality to be defined, 12y + 72 ≥ 0 and 6y − 11 ≥ 0. Solve both inequalities for y.
The domain of y must be restricted to y ≥ . Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (y) = 3(3y − 46)(y − 6). f (y) has real zeros at y =
and y = 6. Set up a sign chart. Substitute a y-value in
each test interval into the factored polynomial to determine if f (y) is positive or negative at that point.
The solutions of 3(3y − 46)(y − 6) ≥ 0 are y-values such that f (y) is positive or equal to 0. From the sign chart, you
can see that the solution set is
. Recall that y ≥ . Create a chart that includes this restriction.
When solving inequalities that involve raising each side to a power to eliminate a radical, it is important to test every
interval using the original inequality. Substitute a y-value in each test interval into the original inequality to determine
if f (y) is a solution.
Thus, the solution set is
43. −
.
< 5
SOLUTION: eSolutions Manual - Powered by Cognero
For the inequality to be defined, 25 − 12x ≥ 0 and 16 − 4x ≥ 0. Solve both inequalities for x.
Page 24
2-6 Thus,
Nonlinear
Inequalities
the solution
set is
43. −
.
< 5
SOLUTION: For the inequality to be defined, 25 − 12x ≥ 0 and 16 − 4x ≥ 0. Solve both inequalities for x.
The domain of x must be restricted to x ≤ . Solve the original inequality.
Make sure to completely isolate the radical before squaring both sides of the inequality. Note change in sign.
Let f (x) = 4(4x − 7)(x + 12). f (x) has real zeros at x =
and x = −12. Set up a sign chart. Substitute an x-value in
each test interval into the factored polynomial to determine if f (x) is positive or negative at that point.
The solutions of 4(4x − 7)(x + 12) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that
the solution set is
. Recall that x ≤ . Create a chart that includes this restriction. When solving
inequalities that involve raising each side to a power to eliminate a radical, it is important to test every interval using
the original inequality. Substitute an x-value in each test interval into the original inequality to determine if f (x) is a
solution.
Notice that x-values that lie outside of the solution set found by solving the inequality for x are also solutions. Thus,
the solution set is
.
Determine the inequality shown in each graph.
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Page 25
Notice that x-values that lie outside of the solution set found by solving the inequality for x are also solutions. Thus,
2-6 the
Nonlinear
solution setInequalities
is
.
Determine the inequality shown in each graph.
44. SOLUTION: Let f (x) be the related function in the graph. f (x) has zeros at x = −2 and x = 5. Thus, (x + 2) and (x − 5) are factors
of f (x). Since the graph has one turning point, f (x) may be quadratic. Start by letting f (x) = (x + 2)(x − 5) or f (x) = x
− 3x − 10. Check using both of the given points, (2, −12) and (4, −6).
2
2
Thus, f (x) = x − 3x − 10. Since the red part of the graph is indicating all points on and above the x-axis, the
2
inequality shown in the graph is x – 3x – 10 ≥ 0.
45. SOLUTION: Let f (x) be the related function in the graph. f (x) has zeros at x = −4 and x = 2. Thus, (x + 4) and (x − 2) are factors
2
of f (x). Since the graph has one turning point, f (x) may be quadratic. Start by letting f (x) = (x + 4)(x − 2) or f (x) = x
+ 2x − 8. Check both of the given points, (−2, −8) and (0, −8).
2
Thus, f (x) = x + 2x − 8. Since the red part of the graph is indicating all points below the x-axis, the inequality shown
2
in the graph is x + 2x – 8 < 0.
Solve each inequality.
46. 2y 4 – 9y 3 – 29y 2 + 60y + 36 > 0
SOLUTION: 4
3
2
Let f (y) = 2y – 9y – 29y + 60y + 36. By using synthetic division, it can be determined that y = 2 is a rational zero.
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By using synthetic division on the depressed polynomial, it can be determined that y = 6 is a rational zero.
Page 26
2
2-6
Thus, f (x) = x + 2x − 8. Since the red part of the graph is indicating all points below the x-axis, the inequality shown
Nonlinear Inequalities
2
in the graph is x + 2x – 8 < 0.
Solve each inequality.
46. 2y 4 – 9y 3 – 29y 2 + 60y + 36 > 0
SOLUTION: 4
3
2
Let f (y) = 2y – 9y – 29y + 60y + 36. By using synthetic division, it can be determined that y = 2 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that y = 6 is a rational zero.
2
The remaining quadratic factor (2y + 7y + 3) can be written as (2y + 1)(y + 3). A factored form of f (y) is f (y) = (y
− 2)(y − 6)(2y + 1)(y + 3). f (y) has real zeros at y = 2, y = 6,
, and y = −3. Set up a sign chart. Substitute a
y-value in each test interval into the factored polynomial to determine if f (y) is positive or negative at that point.
4
3
2
The solutions of 2y – 9y – 29y + 60y + 36 > 0 are y-values such that f (y) is positive. From the sign chart, you can
see that the solution set is
.
47. 3a 4 + 7a 3 – 56a 2 – 80a < 0
SOLUTION: 4
3
2
3
2
Let f (a) = 3a + 7a – 56a – 80a. f (a) can be written as f (a) = a(3a + 7a − 56a − 80). By using synthetic
3
2
division, it can be determined that a = 4 is a rational zero of 3a + 7a − 56a − 80.
2
The remaining quadratic factor (3a + 19a + 20) can be written as (3a + 4)(a + 5). A factored form of f (a) is f (a) =
a(a − 4)(3a + 4)(a + 5). f (a) has real zeros at a = 0, a = 4,
, and a = −5. Set up a sign chart. Substitute an
a-value in each test interval into the factored polynomial to determine if f (a) is positive or negative at that point.
4
3
2
The solutions of 3a + 7a – 56a – 80a < 0 are a-values such that f (a) is negative. From the sign chart, you can
see that the solution set is
.
48. c5 + 6c4 – 12c3 – 56c2 + 96c ≥ 0
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Page 27
SOLUTION: 5
4
3
2
4
3
2
Let f (c) = c + 6c – 12c – 56c + 96c. f (c) can be written as f (c) = c(c + 6c – 12c – 56c + 96). By using
4
3
2
The solutions of 3a + 7a – 56a – 80a < 0 are a-values such that f (a) is negative. From the sign chart, you can
2-6 see
Nonlinear
Inequalities
that the solution
set is
.
48. c5 + 6c4 – 12c3 – 56c2 + 96c ≥ 0
SOLUTION: 5
4
3
2
4
3
2
Let f (c) = c + 6c – 12c – 56c + 96c. f (c) can be written as f (c) = c(c + 6c – 12c – 56c + 96). By using
synthetic division, it can be determined that c = −4 is a rational zero.
By using synthetic division on the depressed polynomial, it can be determined that c = −6 is a rational zero.
2
2
The remaining quadratic factor (c − 4c + 4) can be written as (c − 2) . A factored form of f (c) is f (c) = c(c + 4)(c
2
+ 6)(c − 2) . f (c) has real zeros at c = 0, c = −4, c = −6, and c = 2. Set up a sign chart. Substitute a c-value in each
test interval into the factored polynomial to determine if f (c) is positive or negative at that point.
5
4
3
2
The solutions of c + 6c – 12c – 56c + 96c ≥ 0 are c-values such that f (c) is positive or equal to 0. From the sign
chart, you can see that the solution set is [–6, –4] [0, ).
49. 3x5 + 13x4 – 137x3 – 353x2 + 330x + 144 ≤ 0
SOLUTION: 5
4
3
2
Let f (x) = 3x + 13x – 137x – 353x + 330x + 144. By using synthetic division, it can be determined that x = 1 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = −3 is a rational zero.
By using synthetic division on the new depressed polynomial, it can be determined that x = 6 is a rational zero.
2
The remaining quadratic factor (3x + 25x + 8) can be written as (3x + 1)(x + 8). A factored form of f (x) is f (x) =
(x − 1)(x + 3)(x − 6)(3x + 1)(x + 8). f (x) has real zeros at x = 1, x = −3, x = 6,
, and x = −8. Set up a sign
chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
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5
4
Page 28
3
2
The solutions of 3x + 13x – 137x – 353x + 330x + 144 ≤ 0 are x-values such that f (x) is negative or equal to 0.
5
4
solutions ofInequalities
c + 6c – 12c
2-6 The
Nonlinear
3
2
– 56c + 96c ≥ 0 are c-values such that f (c) is positive or equal to 0. From the sign
chart, you can see that the solution set is [–6, –4] [0, ).
49. 3x5 + 13x4 – 137x3 – 353x2 + 330x + 144 ≤ 0
SOLUTION: 5
4
3
2
Let f (x) = 3x + 13x – 137x – 353x + 330x + 144. By using synthetic division, it can be determined that x = 1 is a
rational zero.
By using synthetic division on the depressed polynomial, it can be determined that x = −3 is a rational zero.
By using synthetic division on the new depressed polynomial, it can be determined that x = 6 is a rational zero.
2
The remaining quadratic factor (3x + 25x + 8) can be written as (3x + 1)(x + 8). A factored form of f (x) is f (x) =
(x − 1)(x + 3)(x − 6)(3x + 1)(x + 8). f (x) has real zeros at x = 1, x = −3, x = 6,
, and x = −8. Set up a sign
chart. Substitute an x-value in each test interval into the factored polynomial to determine if f (x) is positive or
negative at that point.
5
4
3
2
The solutions of 3x + 13x – 137x – 353x + 330x + 144 ≤ 0 are x-values such that f (x) is negative or equal to 0.
From the sign chart, you can see that the solution set is
.
50. PACKAGING A company sells cylindrical oil containers like the one shown.
a. Use the volume of the container to express its surface area as a function of its radius in centimeters. (Hint: 1 liter
= 1000 cubic centimeters)
b. The company wants the surface area of the container to be less than 2400 square centimeters. Write an
inequality that could be used to find the possible radii to meet this requirement.
c. Use a graphing calculator to solve the inequality you wrote in part b and interpret the solution.
SOLUTION: a. If 1 liter is the equivalent of 1000 cubic centimeters, then 2 liters is the equivalent of 2000 cubic centimeters. The
2
formula for the volume V of a cylinder is V = πr h. Substitute 2000 for V and solve for h.
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Page 29
2
inequality that could be used to find the possible radii to meet this requirement.
c. Use a graphing calculator to solve the inequality you wrote in part b and interpret the solution.
SOLUTION: 2-6 a.
Nonlinear
Inequalities
If 1 liter is the
equivalent of 1000 cubic centimeters, then 2 liters is the equivalent of 2000 cubic centimeters. The
2
formula for the volume V of a cylinder is V = πr h. Substitute 2000 for V and solve for h.
2
The formula for the surface area S of a cylinder is S = 2πr + 2πrh. Substitute h =
into the formula for surface area.
2
The surface area as a function of the radius is given by S(r) = 2πr +
.
b. Substitute S(r) = 2400 into the function found in part a and since the company wants the surface area of the
2
container to be less than 2400 square centimeters, 2πr +
< 2400. The inequality can also be written as 2πr
2
− 2400 < 0.
+
2
c. Use a graphing calculator to graph the inequality found in part b. Let y = 2πr +
− 2400. Use the CALC
menu to locate the zeros.
The zeros of the graph are about x = −20.33, x = 1.68, and x = 18.65. The graph is undefined at x = 0. The solutions
2
of 2πr +
− 2400 < 0 are x-values such that f (x) is negative. From the graph, you can see that the solution set
is (–20.33, 0) (1.68, 18.65). Because the radius cannot be negative, the only possible lengths for the radius are between 1.68 centimeters and 18.65 centimeters.
Solve each inequality.
51. (x + 3)2(x – 4)3(2x + 1)2 < 0
SOLUTION: 2
3
2
Let f (x) = (x + 3) (x – 4) (2x + 1) . f (x) has real zeros at x = −3 (multiplicity: 2), x = 4 (multiplicity: 3), and
(multiplicity: 2). Set up a sign chart. Substitute an x-value in each test interval into the polynomial to
determine if f (x) is positive or negative at that point.
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2
3
2
Page 30
The solutions of (x + 3) (x – 4) (2x + 1) < 0 are x-values such that f (x) is negative. Since the inequality cannot
equal 0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is
The zeros of the graph are about x = −20.33, x = 1.68, and x = 18.65. The graph is undefined at x = 0. The solutions
2
− 2400 < 0 are x-values such that f (x) is negative. From the graph, you can see that the solution set
of 2πr +
(–20.33, 0) Inequalities
(1.68, 18.65). Because the radius cannot be negative, the only possible lengths for the radius are 2-6 is
Nonlinear
between 1.68 centimeters and 18.65 centimeters.
Solve each inequality.
51. (x + 3)2(x – 4)3(2x + 1)2 < 0
SOLUTION: 2
3
2
Let f (x) = (x + 3) (x – 4) (2x + 1) . f (x) has real zeros at x = −3 (multiplicity: 2), x = 4 (multiplicity: 3), and
(multiplicity: 2). Set up a sign chart. Substitute an x-value in each test interval into the polynomial to
determine if f (x) is positive or negative at that point.
2
3
2
The solutions of (x + 3) (x – 4) (2x + 1) < 0 are x-values such that f (x) is negative. Since the inequality cannot
equal 0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is
.
52. (y – 5)2(y + 1)(4y – 3)4 ≥ 0
SOLUTION: 2
4
Let f (y) = (y – 5) (y + 1)(4y – 3) . f (y) has real zeros at y = 5 (multiplicity: 2), y = −1, and y =
(multiplicity: 4).
Set up a sign chart. Substitute a y-value in each test interval into the polynomial to determine if f (y) is positive or
negative at that point.
2
4
The solutions of (y – 5) (y + 1)(4y – 3) ≥ 0 are y-values such that f (y) is positive or equal to 0. From the sign chart,
you can see that the solution set is [–1, ).
53. (a – 3)3(a + 2)3(a – 6)2 > 0
SOLUTION: 3
3
2
Let f (a) = (a – 3) (a + 2) (a – 6) . f (a) has real zeros at a = 3 (multiplicity: 3), y = −2 (multiplicity: 3), and a = 6
(multiplicity: 2). Set up a sign chart. Substitute an a-value in each test interval into the polynomial to determine if f (a)
is positive or negative at that point.
3
3
2
The solutions of (a – 3) (a + 2) (a – 6) > 0 are a-values such that f (a) is positive. Since the inequality cannot equal
0, the real zeros are not included in the solution set. From the sign chart, you can see that the solution set is (– , –
2) (3, 6) (6, ).
54. c2(c + 6)3(3c – 4)5(c – 3) ≤ 0
SOLUTION: 2
3
5
Let f (c) = c (c + 6) (3c – 4) (c – 3). f (c) has real zeros at c = 0 (multiplicity: 2), c = −6 (multiplicity: 3), and y =
(multiplicity: 5), and c = 3. Set up a sign chart. Substitute a c-value in each test interval into the polynomial to
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determine if f (c) is positive or negative at that point.
Page 31
3
2-6
3
2
The solutions of (a – 3) (a + 2) (a – 6) > 0 are a-values such that f (a) is positive. Since the inequality cannot equal
0,
the real zerosInequalities
are not included in the solution set. From the sign chart, you can see that the solution set is (– , –
Nonlinear
2) (3, 6) (6, ).
54. c2(c + 6)3(3c – 4)5(c – 3) ≤ 0
SOLUTION: 2
3
5
Let f (c) = c (c + 6) (3c – 4) (c – 3). f (c) has real zeros at c = 0 (multiplicity: 2), c = −6 (multiplicity: 3), and y =
(multiplicity: 5), and c = 3. Set up a sign chart. Substitute a c-value in each test interval into the polynomial to
determine if f (c) is positive or negative at that point.
2
3
5
The solutions of c (c + 6) (3c – 4) (c – 3) ≤ 0 are c-values such that f (c) is negative or equal to 0. From the sign
chart, you can see that the solution set is
.
55. STUDY TIME Jarrick determines that with the information that he currently knows, he can achieve a score of a 75% on his test. Jarrick believes that for every 5 complete minutes he spends studying, he will raise his score by 1%.
a. If Jarrick wants to obtain a score of at least 89.5%, write an inequality that could be used to find the time t that he
will have to spend studying.
b. Solve the inequality that you wrote in part a and interpret the solution.
SOLUTION: a. Without studying, Jarrick starts at a score of 75%. For every 5 complete minutes that he spends studying, 1% can
be added to his score. Since fractions of 5 minutes will not increase his score, the greatest integer function is used to
find out how many times he spends 5 complete minutes studying out of a total of t minutes. Thus, his score S after
studying for t minutes can be found using
If Jarrick wants to obtain a score of at least 89.5%, the equation becomes the following inequality:
b. Sample answer:
Since t ≥ 72.5, Jarrick will have to spend 75 minutes studying for the test.
56. GAMES A skee ball machine pays out 3 tickets each time a person plays and then 2 additional tickets for every 80 points the player scores.
a. Write a nonlinear function to model the amount of tickets received for an x-point score.
b. Write an inequality that could be used to find the score a player would need in order to receive at least 11 tickets
c. Solve the inequality in part b and interpret your solution.
SOLUTION: a. For every 80 points a player scores, the machine will pay out 2 tickets. Write an expression that represents this
situation using the greatest integer function.
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Since the machine pays out 3 tickets just for playing, the function becomes
Page 32
.
2-6 Nonlinear Inequalities
Since t ≥ 72.5, Jarrick will have to spend 75 minutes studying for the test.
56. GAMES A skee ball machine pays out 3 tickets each time a person plays and then 2 additional tickets for every 80 points the player scores.
a. Write a nonlinear function to model the amount of tickets received for an x-point score.
b. Write an inequality that could be used to find the score a player would need in order to receive at least 11 tickets
c. Solve the inequality in part b and interpret your solution.
SOLUTION: a. For every 80 points a player scores, the machine will pay out 2 tickets. Write an expression that represents this
situation using the greatest integer function.
Since the machine pays out 3 tickets just for playing, the function becomes
.
b. Substitute f (x) = 11 and written the function found in part a as an inequality to represent the situation.
c.
The solution set is [320,
tickets.
). So, any score at or above 320 points will result in the player receiving at least 11
57. The area of a region bounded by a parabola and a horizontal line is A =
bh, where b represents the base of the
region along the horizontal line and h represents the height of the region. Find the area bounded by f and g.
SOLUTION: b is 5 − (−1) or 6. To find h, find f (x).
f(x) has zeros at x = −2 and x = 6. Thus, (x + 2) and (x − 6) are factors of f (x). Since the graph has one turning
2
point, f (x) may be quadratic. Start by letting f (x) = (x + 2)(x − 6) or f (x) = x − 4x − 12. Check using both of the
given points, (−1, −7) and (5, −7).
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2
Page 33
solution setInequalities
is [320, ). So, any score at or above 320 points will result in the player receiving at least 11
2-6 The
Nonlinear
tickets.
57. The area of a region bounded by a parabola and a horizontal line is A =
bh, where b represents the base of the
region along the horizontal line and h represents the height of the region. Find the area bounded by f and g.
SOLUTION: b is 5 − (−1) or 6. To find h, find f (x).
f(x) has zeros at x = −2 and x = 6. Thus, (x + 2) and (x − 6) are factors of f (x). Since the graph has one turning
2
point, f (x) may be quadratic. Start by letting f (x) = (x + 2)(x − 6) or f (x) = x − 4x − 12. Check using both of the
given points, (−1, −7) and (5, −7).
2
Thus, f (x) = x − 4x − 12. The vertex is of f (x) is located at the point with the x-coordinate,
.
Substitute x = 2 into f (x).
The vertex of f (x) is at (2, −16). The distance from the vertex to the horizontal line is the height of the region. Thus,
h is
or 9.
Substitute h = 9 and b = 6 into A =
bh.
The area bounded by f and g is 36 square units.
If k is nonnegative, find the interval for x for which each inequality is true.
58. x2 + k x + c ≥ c
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SOLUTION: 2
x + k x + c ≥ c can be written as
Page 34
2-6 Nonlinear Inequalities
The area bounded by f and g is 36 square units.
If k is nonnegative, find the interval for x for which each inequality is true.
58. x2 + k x + c ≥ c
SOLUTION: 2
x + k x + c ≥ c can be written as
Let f (x) = x(x + k). The zeros of f (x) are at x = 0 and x = −k. Set up a sign chart. Because the degree of f is even
and its leading coefficient is positive,
This means that the function starts off positive
at the left and ends positive at the right. Because each zero listed is the location of a sign change, the chart can be
completed as shown.
2
The solutions of x + k x + c ≥ c are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
that the solution set is x ≤ −k or x ≥ 0. Thus, (– , –k] [0, ).
59. (x + k)(x – k) < 0
SOLUTION: Let f (x) = (x + k)(x – k). The zeros of f (x) are at x = −k and x = k. Set up a sign chart. Because the degree of f is
even and its leading coefficient is positive,
This means that the function starts off positive at the left and ends positive at the right. Because each zero listed is the location of a sign change, the chart
can be completed as shown.
The solutions of (x + k)(x – k) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the
solution set is –k < x < k. Thus, (–k, k).
60. x3 – k x2 – k 2x + k 3 > 0
SOLUTION: 3
2
2
3
x – k x – k x + k > 0 can be written as
2
Let f (x) = (x + k)(x − k) . The zeros of f (x) are at x = −k and x = k (multiplicity: 2). Set up a sign chart. Because
the degree of f is odd and its leading coefficient is positive,
This means that the function starts off negative at the left and ends positive at the right. The zero at x = −k is the location of a sign
change
because
it has
multiplicity 1. The zero at x = k is not the location of a sign change because it has multiplicity
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2.
solutions of
(x + k)(x – k) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the
2-6 The
Nonlinear
Inequalities
solution set is –k < x < k. Thus, (–k, k).
60. x3 – k x2 – k 2x + k 3 > 0
SOLUTION: 3
2
2
3
x – k x – k x + k > 0 can be written as
2
Let f (x) = (x + k)(x − k) . The zeros of f (x) are at x = −k and x = k (multiplicity: 2). Set up a sign chart. Because
the degree of f is odd and its leading coefficient is positive,
This means that the function starts off negative at the left and ends positive at the right. The zero at x = −k is the location of a sign
change because it has multiplicity 1. The zero at x = k is not the location of a sign change because it has multiplicity
2.
3
2
2
3
The solutions of x – k x – k x + k > 0 are x-values such that f (x) is positive. From the sign chart, you can see that
the solution set is x > −k. Thus, (−k, ).
61. x4 – 8k 2x2 + 16k 4 ≥ 0
SOLUTION: 4
2 2
4
x – 8k x + 16k ≥ 0 can be written as
2
2
Let f (x) = (x − 2k) (x + 2k) . The zeros of f (x) are at x = 2k (multiplicity: 2) and x = −2k (multiplicity: 2). Set up a
sign chart. Because the degree of f is even and its leading coefficient is positive,
This means that the function starts off positive at the left and ends positive at the right. Because each zero listed has multiplicity 2, neither is the location of a sign change.
4
2 2
4
The solutions of x – 8k x + 16k ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you
can see that the solution set is all read numbers. Thus, (– , ).
62. MULTIPLE REPRESENTATIONS In this problem, you will investigate absolute value nonlinear inequalities.
a. TABULAR Copy and complete the table below.
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Page 36
4
2 2
4
The solutions of x – 8k x + 16k ≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you
can see that the solution set is all read numbers. Thus, (– , ).
2-6
Nonlinear Inequalities
62. MULTIPLE REPRESENTATIONS In this problem, you will investigate absolute value nonlinear inequalities.
a. TABULAR Copy and complete the table below.
b.GRAPHICAL Graph each function in part a.
c. SYMBOLIC Create a sign chart for each inequality given. Include both zeros and undefined points and evaluate
the sign of the numerators and denominators separately.
i.
ii.
iii.
d. NUMERICAL Write the solution for each inequality in part c.
SOLUTION: a. The zero of f (x) occurs when the expression in the numerator, x − 1, is equal to 0. Thus, x = 1 is a zero of f (x). f
(x) is undefined when the expression in the denominator,
, is equal to 0. Thus, f (x) is undefined at x = −2.
The zero of g(x) occurs when the expression in the numerator,
, is equal to 0. Thus, x =
is a zero of f (x). f
(x) is undefined when the expression in the denominator, x − 3, is equal to 0. Thus, f (x) is undefined at x = 3.
The zero of h(x) occurs when the expression in the numerator,
, is equal to 0. Thus, x = −4 is a zero of f (x). f
(x) is undefined when the expression in the denominator,
, is equal to 0. Thus, f (x) is undefined at x =
.
b. Use a graphing calculator to graph each function.
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c. i.
Page 37
2-6 Nonlinear Inequalities
b. Use a graphing calculator to graph each function.
c. i.
ii.
iii.
d. i. f (x) < 0 on (−
, −2)
(−2, 1)
ii. g(x) ≥ 0 on
iii.
63. ERROR ANALYSIS Ajay and Mae are solving
and Mae thinks that the solution is (−
,
. Ajay thinks that the solution is (−
, 0] or [0,
),
). Is either of them correct? Explain your reasoning.
SOLUTION: eSolutions Manual - Powered by Cognero
Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator, x = 0
(multiplicity: 2), and denominator, x = 3. Create a sign chart. Then choose and test x-values in each interval to
Page 38
ii. g(x) ≥ 0 on
2-6 iii.
Nonlinear Inequalities
63. ERROR ANALYSIS Ajay and Mae are solving
and Mae thinks that the solution is (−
,
. Ajay thinks that the solution is (−
, 0] or [0,
),
). Is either of them correct? Explain your reasoning.
SOLUTION: Let f (x) =
. The zeros and undefined points of the inequality are the zeros of the numerator, x = 0
(multiplicity: 2), and denominator, x = 3. Create a sign chart. Then choose and test x-values in each interval to
determine if f (x) is positive or negative.
≥ 0 are x-values such that f (x) is positive or equal to 0. From the sign chart, you can see
The solutions of
that the solution set is (– , 3) (3, ). x = 3 is not part of the solution set because the original inequality is
undefined at x =3. Thus, neither Ajay nor Mae is correct.
64. REASONING If the solution set of a polynomial inequality is (−3, 3), what will be the solution set if the inequality
symbol is reversed? Explain your reasoning.
SOLUTION: If the inequality symbol is reversed, the solution set will consist of all real numbers not in the original solution set.
Thus, the new solution set will be (– , –3) (3, ).
65. CHALLENGE Determine the values for which (a + b)2 > (c + d)2 if a < b < c < d.
SOLUTION: If a > 0, then b, c, and d are all positive and
and (c + d) are squared, |a + b| > |c + d|.
. Thus, a < 0. Also, before the expressions (a + b)
66. REASONING If 0 < c < d, find the interval on which (x – c)(x – d) ≤ 0 is true. Explain your reasoning.
SOLUTION: Because c < d, when x < c, x < d, so (x – c) and (x – d) will be negative and (x – c)(x – d) will be positive. When x
> d, both factors will be positive, so (x – c)(x – d) will be positive. When c ≤ x ≤ d, (x – c) will be positive or zero,
(x – d) will be negative or zero, and (x – c)(x – d) will be either negative or zero. Thus, the solution set is [c, d].
67. CHALLENGE What is the solution set of (x − a)2n > 0 if n is a natural number?
SOLUTION: 2n
Because (x − a) is always an even degree, by definition, the end behavior is to approach ∞ as x → ±∞. The absolute minimum occurs as x − a = 0 or x = a. Since any other value of (x − a) to an even power must be positive
2n
2n
in the reals, (x − a) > 0 for all values of x except when x = a, since (0) > 0 is false. Thus, the solution in set is
(− , a) (a, ).
68. REASONING What happens to the solution set of (x + a)(x − b) < 0 if the expression is changed to −(x + a)(x −
b) < 0, where a and b > 0? Explain your reasoning.
SOLUTION: Sample answer: Let f (x) = (x + a)(x − b). The zeros of f (x) are at x = −a and x = b. Set up a sign chart. Because
Page 39
leading coefficient is positive,
This means that the eSolutions
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the degree
f is even
and its
function starts off positive at the left and ends positive at the right. Because each zero listed is the location of a sign
change, the chart can be completed as shown.
2n
2-6
Because (x − a) is always an even degree, by definition, the end behavior is to approach ∞ as x → ±∞. The absolute minimum occurs as x − a = 0 or x = a. Since any other value of (x − a) to an even power must be positive
2n
2n
in
the reals, (x Inequalities
− a) > 0 for all values of x except when x = a, since (0) > 0 is false. Thus, the solution in set is
Nonlinear
(− , a) (a, ).
68. REASONING What happens to the solution set of (x + a)(x − b) < 0 if the expression is changed to −(x + a)(x −
b) < 0, where a and b > 0? Explain your reasoning.
SOLUTION: Sample answer: Let f (x) = (x + a)(x − b). The zeros of f (x) are at x = −a and x = b. Set up a sign chart. Because
the degree of f is even and its leading coefficient is positive,
This means that the function starts off positive at the left and ends positive at the right. Because each zero listed is the location of a sign
change, the chart can be completed as shown.
The solutions of (x + a)(x − b) < 0 are x-values such that f (x) is negative. From the sign chart, you can see that the
solution set is (–a, b). If the left side of the expression is multiplied by −1, the zeros will remain the same but the sign
of the expression will change for each interval.
The solution set for the new inequality is (− , −a) (b, ). Thus, the zeros will remain the same but the solution
set will consist of all numbers outside of the original solution set.
69. Writing in Math Explain why you cannot solve
< 6 by multiplying each side by x – 2.
SOLUTION: Sample answer: There will be instances when x – 2 is a negative value. When this occurs, we are multiplying an
inequality by a negative and the inequality symbol will need to be reversed. The problem occurs here because x – 2
could be positive or negative.
Find the domain of each function and the equations of any vertical or horizontal asymptotes, if any.
70. f (x) =
SOLUTION: The function is undefined at the real zeros of the denominator b(x) = x + 4. The real zero of b(x) is −4. Therefore, D
= {x | x –4, x R}.
Check for vertical asymptotes.
Determine whether x = −4 is a point of infinite discontinuity. Find the limit as x approaches −4 from the left and the
right.
x −4.1 −4.01 −4.001 −4 −3.999 −3.99 −3.9
f(x) 82 802 8002 undef −7998 −798 −78
Because
x = −4 is a vertical asymptote of f .
Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).
x
0
50
100
1000
−1000 −100
−50
f (x)
2.004
2.08
2.17
0
1.85
1.92
1.99
The table suggests
Therefore, you know that y = 2 is a horizontal asymptote of f .
71. h(x) =
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Page 40
The function is undefined at the real zeros of the denominator b(x) = x + 6. The real zero of b(x) is −6. Therefore, D
= {x | x –6, x R}.
2-6
Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).
x
0
50
100
1000
−1000 −100
−50
f (x)
2.004
2.08
2.17
0
1.85
1.92
1.99
The
table suggests
Therefore, you know that y = 2 is a horizontal asymptote of f .
Nonlinear
Inequalities
71. h(x) =
SOLUTION: The function is undefined at the real zeros of the denominator b(x) = x + 6. The real zero of b(x) is −6. Therefore, D
= {x | x –6, x R}.
Check for vertical asymptotes.
Determine whether x = −6 is a point of infinite discontinuity. Find the limit as x approaches −6 from the left and the
right.
x −6.1 −6.01 −6.001 −6 −5.999 −5.99 −5.9
h
−372 −3612 −36,012 undef 35,988 3588 348
(x)
Because
x = −6 is a vertical asymptote of f .
Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).
x
0
10
50
100
−100
−50
−10
f (x) −106.4 −56.8
0
6.25
44.6
94.3
−25
The table suggests
Therefore, there is no horizontal asymptote.
72. f (x) =
SOLUTION: The function is undefined at the real zeros of the denominator b(x) = (2x + 1)(x − 5). The real zeros of b(x)
are
. Therefore,
.
Check for vertical asymptotes.
Determine whether
the right.
x
f (x)
−0.51
−13.7
is a point of infinite discontinuity. Find the limit as x approaches
−0.501 −0.5001 −0.5 −0.4999 −0.499
136
−136 −1364 undef 1364
Because
from the left and −0.49
13.6
is a vertical asymptote of f .
Determine whether x = 5 is a point of infinite discontinuity. Find the limit as x approaches 5 from the left and the
right.
x
4.9
4.99
4.999
5
5.001
5.01
5.1
f (x)
364
36.4
3.7
−3.6
−36.3 −364 undef
Because
x = 5 is a vertical asymptote of f .
Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).
x
0
10
50
100
−100
−50
−10
f (x) −0.005 −0.009 −0.039
0.2
0.086 0.011 0.005
The table suggests
Therefore, you know that y = 0 is a horizontal asymptote of f .
73. GEOMETRY A cone is inscribed in a sphere with a radius of 15 centimeters. If the volume of the cone is 1152π
cubic centimeters, find the length represented by x.
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Page 41
2-6
Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).
x
0
10
50
100
−100
−50
−10
f (x) −0.005 −0.009 −0.039
0.2
0.086 0.011 0.005
Nonlinear
Inequalities
The table suggests
Therefore, you know that y = 0 is a horizontal asymptote of f .
73. GEOMETRY A cone is inscribed in a sphere with a radius of 15 centimeters. If the volume of the cone is 1152π
cubic centimeters, find the length represented by x.
SOLUTION: The formula for the volume of a cone is V =
2
πr h. The height of the cone, h, is equal to 15 + x. Substitute V =
1152π and h = 15 + x into the formula.
2
The triangle formed by x, the radius of the cone (r), and the radius of the sphere (15) is a right triangle. Thus, x + r
2
2
2
= 15 or 225 − x = r .
2
2
2
2
Substitute r = 225 − x into 3456 = r (15 + x) and solve for x.
3
2
Let f (x) = x + 15x − 225x + 81. By using synthetic division on the depressed polynomial, it can be determined that
x = 9 is a rational zero.
2
The remaining quadratic factor (x + 24x − 9) yields no rational zeros. Use the quadratic formula to find the zeros.
The zeros of f (x) are x = 9, x = −12 + 3
of x must be positive, this eliminates −12 − 3
or about 0.37, and x = −12 − 3
or about −24.37. Since the length
. Thus, x = 9 cm or about 0.37 cm.
Divide using long division.
74. (x2 – 10x – 24) ÷ (x + 2)
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Page 42
The zeros of f (x) are x = 9, x = −12 + 3
or about 0.37, and x = −12 − 3
2-6 Nonlinear Inequalities
of x must be positive, this eliminates −12 − 3
or about −24.37. Since the length
. Thus, x = 9 cm or about 0.37 cm.
Divide using long division.
74. (x2 – 10x – 24) ÷ (x + 2)
SOLUTION: 2
So, (x – 10x – 24) ÷ (x + 2) = x – 12.
75. (3a 4 – 6a 3 – 2a 2 + a – 6) ÷ (a + 1)
SOLUTION: 4
3
2
3
2
So, (3a – 6a – 2a + a – 6) ÷ (a + 1) = 3a – 9a + 7a – 6.
76. (z 5 – 3z 2 – 20) ÷ (z – 2)
SOLUTION: 5
2
So, (z
– 3z- Powered
– 20) by
÷ (z
– 2) = z
eSolutions
Manual
Cognero
77. (x3 + y 3) ÷ (x + y)
4
3
2
+ 2z + 4z + 5z + 10.
Page 43
2-6 Nonlinear
4
3 Inequalities
2
3
2
So, (3a – 6a – 2a + a – 6) ÷ (a + 1) = 3a – 9a + 7a – 6.
76. (z 5 – 3z 2 – 20) ÷ (z – 2)
SOLUTION: 5
2
4
3
2
So, (z – 3z – 20) ÷ (z – 2) = z + 2z + 4z + 5z + 10.
77. (x3 + y 3) ÷ (x + y)
SOLUTION: 3
3
2
2
So, (x + y ) ÷ (x + y) = x – xy + y .
78. FINANCE The closing prices in dollars for a share of stock during a one-month period are shown.
a. Graph the data.
b. Use a graphing calculator to model the data using a polynomial function with a degree of 3.
c. Use the model to estimate the closing price of the stock on day 25.
SOLUTION: a. Use a graphing calculator to plot the data.
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2-6 Nonlinear
3
3 Inequalities
2
2
So, (x + y ) ÷ (x + y) = x – xy + y .
78. FINANCE The closing prices in dollars for a share of stock during a one-month period are shown.
a. Graph the data.
b. Use a graphing calculator to model the data using a polynomial function with a degree of 3.
c. Use the model to estimate the closing price of the stock on day 25.
SOLUTION: a. Use a graphing calculator to plot the data.
b. Use the cubic regression function on the graphing calculator.
3
2
f(x) = 0.003x − 0.111x + 0.019x + 30.259
c. Graph the regression equation using a graphing calculator. Use the CALC function on the graphing calculator to
find the price of a share of stock on Day 25.
The price of the stock on Day 25 was about $8.42.
79. HOME SECURITY A company offers a home-security system that uses the numbers 0 through 9, inclusive, for a
5-digit security code.
a. How many different security codes are possible?
b. If no digits can be repeated, how many security codes are available?
c. Suppose the homeowner does not want to use 0 or 9 as the first digit and wants the last digit to be 1. How many
codes can be formed if the digits can be repeated? If no repetitions are allowed, how many codes are available?
SOLUTION: a. There are 5 digits for the code with 10 possibilities for each digit. Thus, there are 10 · 10 · 10 · 10 · 10 or 100,000 possible security codes.
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b. The
order
of the by
numbers
2-6 Nonlinear Inequalities
The price of the stock on Day 25 was about $8.42.
79. HOME SECURITY A company offers a home-security system that uses the numbers 0 through 9, inclusive, for a
5-digit security code.
a. How many different security codes are possible?
b. If no digits can be repeated, how many security codes are available?
c. Suppose the homeowner does not want to use 0 or 9 as the first digit and wants the last digit to be 1. How many
codes can be formed if the digits can be repeated? If no repetitions are allowed, how many codes are available?
SOLUTION: a. There are 5 digits for the code with 10 possibilities for each digit. Thus, there are 10 · 10 · 10 · 10 · 10 or 100,000 possible security codes.
b. The order of the numbers in the code is important, so this situation is a permutation of 10 digits taken 5 at a time.
Thus, there are 30,240 possible security codes.
c. If the first digit cannot be 0 or 9, then there are 8 possibilities. Since digits can be repeated, there are 10
possibilities for the second, third, and fourth digits. There is only one possibility for the last digit, 1. Thus, there are 8 ·
10 · 10 · 10 · 1 or 8000 possible codes if the digits can be repeated.
If the first digit cannot be 0 or 9, then there are 7 possibilities (remember, 1 has to be the last digit, and thus, cannot
be the first digit). Since digits cannot be repeated, there are just 8 possibilities for the second digit (both the first digit
and the last digit, 1, are excluded). There are then 7 possibilities for the third digit (the first two digits and the last
digit, 1, are excluded), 6 possibilities for the fourth, and 1 possibility for the fifth digit. Thus, there are 7 · 8 · 7 · 6 · 1 or 2352 possible codes if no repetitions are allowed.
80. SAT/ACT Two circles, A and B, lie in the same plane. If the center of circle B lies on circle A, then in how many
points could circle A and circle B intersect?
I. 0
II. 1
III. 2
A I only
B III only
C I and III only
D II and III only
E I, II, and III
SOLUTION: Draw all possibilities.
The correct answer is E.
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81. A rectangle is 6 centimeters longer than it is wide. Find the possible widths if the area of the rectangle is more than
2-6
If the first digit cannot be 0 or 9, then there are 7 possibilities (remember, 1 has to be the last digit, and thus, cannot
be the first digit). Since digits cannot be repeated, there are just 8 possibilities for the second digit (both the first digit
and the last digit, 1, are excluded). There are then 7 possibilities for the third digit (the first two digits and the last
digit, 1, are excluded), 6 possibilities for the fourth, and 1 possibility for the fifth digit. Thus, there are 7 · 8 · 7 · 6 · 1 Nonlinear Inequalities
or 2352 possible codes if no repetitions are allowed.
80. SAT/ACT Two circles, A and B, lie in the same plane. If the center of circle B lies on circle A, then in how many
points could circle A and circle B intersect?
I. 0
II. 1
III. 2
A I only
B III only
C I and III only
D II and III only
E I, II, and III
SOLUTION: Draw all possibilities.
The correct answer is E.
81. A rectangle is 6 centimeters longer than it is wide. Find the possible widths if the area of the rectangle is more than
216 square centimeters.
F w > 12
G w < 12
H w > 18
J w < 18
SOLUTION: The formula for the area of a rectangle is A = lw. An expression for the length of the rectangle is l = w + 6. Use A =
216 and l = w + 6 to write an inequality that represents the situation.
Let f (x) = (w + 18)(w − 12). f (x) has real zeros at x = −18 and x = 12. Set up a sign chart. Substitute an x-value in
each test interval into the polynomial to determine if f (x) is positive or negative at that point.
The solutions of (x + 18)(x − 12) > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the
solution set is (– , –18) (12, ). Since the width must be positive, the solution set is (12, ) or w > 12. The
correct answer is F.
82. FREE RESPONSE The amount of drinking water reserves in millions of gallons available for a town is modeled
2
by f (t) = 80 + 10t − 4t . The minimum amount of water needed by the residents is modeled by
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is the
time-in
years.by Cognero
a. Identify the types of functions represented by f (t) and g(t).
b. What is the relevant domain and range for f (t) and g(t)? Explain.
where t
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2-6
The solutions of (x + 18)(x − 12) > 0 are x-values such that f (x) is positive. From the sign chart, you can see that the
solution set is (– , –18) (12, ). Since the width must be positive, the solution set is (12, ) or w > 12. The
correct
answerInequalities
is F.
Nonlinear
82. FREE RESPONSE The amount of drinking water reserves in millions of gallons available for a town is modeled
2
by f (t) = 80 + 10t − 4t . The minimum amount of water needed by the residents is modeled by
where t
is the time in years.
a. Identify the types of functions represented by f (t) and g(t).
b. What is the relevant domain and range for f (t) and g(t)? Explain.
c. What is the end behavior of f (t) and g(t)?
d. Sketch f (t) and g(t) for 0 ≤ t ≤ 6 on the same graph.
e . Explain why there must be a value c for [0,6] such that f (c) = 50.
f. For what value in the relevant domain does f have a zero? What is the significance of the zero in this situation?
g. If this was a true situation and these projections were accurate, when would the residents be expected to need
more water than they have in reserves?
SOLUTION: a. f (t) is a quadratic polynomial function. g(t) is a power or radical function.
b. Since time and the amount of water cannot be negative, the relevant domains for f (t) and g(t) are restricted to
nonnegative values for t that result in nonnegative values for f (t) and g(t). Graph both functions using a graphing
calculator. For f (t), use the CALC function to locate the zero.
The domain for f (t) is D = [0, 5.89]. For the range, find the maximum using
The vertex occurs when t =
.
. Find f (t).
The range for f (t) is R = [0, 86.25].
For g(t), D = [0, ∞) and R = [0, ∞).
c. Use a graphing calculator to graph both functions. Since the domain of f (t) is D = [0, 5.89], analyze f (t) as t
approaches both 0 and 5.89. Since the domain of g(t) is D = [0, ∞), analyze g(t) as t approaches both 0 and ∞.
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2-6
The range for f (t) is R = [0, 86.25].
For g(t), D = [0, ∞) and R = [0, ∞).
Nonlinear Inequalities
c. Use a graphing calculator to graph both functions. Since the domain of f (t) is D = [0, 5.89], analyze f (t) as t
approaches both 0 and 5.89. Since the domain of g(t) is D = [0, ∞), analyze g(t) as t approaches both 0 and ∞.
For f (t), as t approaches 0, f (t) approaches 80. As t approaches 5.89, f (t) approaches 0. For g(t), as t approaches 0,
g(t) approaches 0. As t approaches ∞, g(t) approaches ∞.
d. Evaluate the functions for several t-values for 0 ≤ t ≤ 6.
e . f (x) is a polynomial function, so it is also continuous and the Intermediate Value Theorem applies. Therefore,
since f (3) = 74 and f (5) = 30, it follows that there is a number c, such that 3 < c < 5 and f (c) = 50.
f. The domain of f (t) is D = [0, 5.89]. When t ≈ 5.89, f (t) = 0. This means that the town will run out of water
reserves after about 5.89 years.
e . Use a graphing calculator to find the intersection of the two graphs.
When t = 5.23, the two graphs intersect. When t > 5.23, g(t) > f (t). In about 5.23 years, the residents will need more
water than what is in their reserves.
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