MT100.22 ASSIGNMENT ONE—SOLUTIONS
n
o
x
1.1.24 Describe the set x : x+1
< 0 as an interval.
x
is
Solution. The problem asks us to describe the set of x such that the quotient x+1
negative. A quotient may be negative only when the numerator and denominator have
different signs.
We must consider, therefore, the two separate situations
(1)
x<0
and
x+1 >0
x>0
and
x+1 <0
and
(2)
The right condition in situation (1) says −1 < x. Combining this with left condtion, we get
that −1 < x < 0 must be the case.
Looking at the right condition in situation (2), we see that x < −1 must be true. Of
course, the left condition says that x > 0. Since there are no real numbers which are
simultaneously smaller than −1 and
n larger than
o 0, this situation is impossible.
x
We conclude then that the set x : x+1 < 0 is described by x : −1 < x < 0, which is in
turn the interval (−1, 0).
1.2.10 Find the equation of the line with slope −2 and y-intercept 3.
Solution. Remembering that slope-intercept form of the line is y = mx + b where m is
the slope and b is the y-intercept, an equation for this line is
y = −2x + 3
1.2.24 A line of slope m = 2 passes through (1, 4). Find y such that (3, y) lies on the line.
Solution. Here are a couple of ways to proceed:
(1) Notice that the slope is defined as
y2 − y 1
m=
x2 − x 1
for any two points (x1 , y1 ) and (x2 , y2 ) on the line. It follows that
y−4
2=
3−1
Simplifying and solving for y gives that y = 8.
(2) Using point-slope form of a line, we get that this line has equation
y − 4 = 2(x − 1)
Plugging 3 in for x, we get that y = 8.
1.2.26 Assume that the number N of concert tickets which can be sold at a price of P dollars per
ticket is a linear function N (P ) for 10 ≤ P ≤ 40. Determine N (P ) (called the demand
function) if N (10) = 500 andN (40) = 0. What is the decrease ∆N in the number of tickets
sold if the price is increased by ∆P = 5 dollars?
Solution. Sentence number one tells us that the number of tickets is expressed in terms
of the price as
N (P ) = mP + b
1
Also, we know that the points (10, 500) and (40, 0) are on that line, and so
−50
500 − 0
=
10 − 40
3
m=
Using the point (40, 0) and point-slope form of the line, we get that
N −0=
−50
(P − 40)
3
This is the demand function.
Now, as number of tickets sold and price are linearly related, the slope of the line m is
equal to ∆N
∆P . So, when ∆P = 5, we get that
−50
−250
(5) =
= −83.3
3
3
So, when the price is increased by $5, we can expect 83 and one-third fewer tickets to be
demanded.
∆N = m∆P =
1.2.40 Complete the square and find the minimum or maximum value of the quadratic function
y = 3x2 + 12x − 5.
Solution. Begin by factoring out the 3 to get y = 3(x 2 +4x)−5. Now, divide the 4 by two
and square. Add and subtract that value inside the parentheses: y = 3(x 2 + 4x + 4 − 4) − 5
Now rewrite this as y = 3(x2 + 4x + 4) − 12 − 5. Simplify this to get y = 3(x + 2) 2 − 17.
This equation tells us that the parabola opens upward (since the coefficient 3 is a positive
number), and therefore the vertex is a minimum. This equation also tells us that the vertex
is (−2, −17). Therfore, the minimum value is −17.
1.2.50 Find the numbers x and y with sum 10 and product 24. Hint: Find a quadratic polynomial
satisfied by x.
Solution. We need to find x and y satisfying
x + y = 10
xy = 24
Solving the second of these equation for y and substituting into the first one, we obtain
24
= 10.
x
As x must not be zero (we know this because the product of x and y is not zero), we can
multiply through to get the equation
x+
x2 + 24 = 10x
which is equivalent to
x2 − 10x + 24 = 0
You can factor, or use the quadratic formula to get
√
10 ± 100 − 96
x=
= 6, 4
2
When x = 6, y = 4, and when x = 4, y = 6. In either case, the two numbers are 4 and 6.
1.3.26 Show that f (x) = x2 + 3x−1 and g(x) = 3x3 − 9x + x−2 are rational functions (show that
each is a quotient of polynomials).
Solution. Notice that f (x) = x2 + x3 . The common denominator here is x, so
f (x) =
3
x3 + 3
x3
+ =
x
x
x
2
A similar thing may be done with g; in this case the common denominator is x 2 :
1
g(x) = 3x3 − 9x + 2
x
3x5 9x3
1
= 2 − 2 + 2
x
x
x
3x5 − 9x3 + 1
=
x2
1.3.36 Find all values of c such that the domain of
x+1
f (x) = 2
x + 2cx + 4
is R.
Solution. Every real number is in the domain of f if c is a number such that the denominator has no real roots. The roots of the denominator are given by
√
−2c ± 4c2 − 16
x=
2
There are no real values for these roots when 4c 2 − 16 < 0. This implies that c2 < 4 and
therefore, the domain of f is all real numbers if −2 < c < 2.
1.4.12 Find all the angles between 0 and 2π satisfying csc(θ) = 2.
1
, we must find all of the angles between 0 and 2π
Solution. Recalling that csc(θ) = sin(θ)
1
such that sin(θ) = 2 . This occurs for angles in the first and second quadrant. The angles
π
5π
whose sine is 21 are π6 and 5π
6 . Thus the angles whose cosecant is 2 are 6 and 6 .
p
1.4.24 Find sin(2θ) and cos(2θ) if tan(θ) = (2).
√
sin(θ)
Solution. Remember that tan(θ) = cos(θ)
. If tan(θ) = 2, then we have sin(θ) =
√
2 cos(θ). Using the Pythagorean identity for sine and cosine and substituting, we have
1 = sin2 (θ) + cos2 (θ) = 2 cos2 (θ) + cos2 (θ) = 3 cos 2 (θ)
√
It follows that cos(θ) = √13 and sin(θ) = √23 .
To compute sin(2θ) and cos(2θ), use the double angle formulas:
√
√
2 1
2 2
sin(2θ) = 2 sin(θ) cos(θ) = 2 √ √ =
3
3 3
−1
1 2
cos(2θ) = cos2 (θ) − sin2 (θ) = − =
3 3
3
1.4.44 Show that
sin(θ + π) = − sin(θ)
Solution. Use the sum of angles formula for sine to see that
sin(θ + π) = sin(θ) cos(π) + cos(θ) sin(π) = − sin(θ)
since cos(π) = −1 and sin(π) = 0.
3