Basic Mathematical Skills
Exercises
Version: August 24th, 2017
last changes in purple colour
Contents
1 Expressions with brackets
1.1 Expanding brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Special products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Factorizing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
3
3
2 Fractions
2.1 Computing with fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Fractions with variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Solving equations with fractions . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
6
6
3 Square roots
8
4 Exponents and logarithms
4.1 Computing with exponents . . . . .
4.2 Roots as exponents . . . . . . . . .
4.3 Solving exponential equations . . .
4.4 Logarithms: applying the definition
4.5 Logarithms: applying the rules . .
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9
9
9
10
10
11
5 Trigonometric functions
5.1 Relations between sine and cosine
5.2 Exact values . . . . . . . . . . . .
5.3 Inverse trigonometric functions .
5.4 Addition formulas . . . . . . . . .
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12
12
13
14
14
6 Solving equations
6.1 Linear equations . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Quadratic equations . . . . . . . . . . . . . . . . . . . . . . .
6.2.1 Factorizing . . . . . . . . . . . . . . . . . . . . . . . .
6.2.2 The quadratic formula (also known as the abc-formula)
6.3 Equations of higher degree . . . . . . . . . . . . . . . . . . . .
6.4 A2 = B 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.5 √
A·B =A·C . . . . . . . . . . . . . . . . . . . . . . . . . . .
A=B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.6
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16
16
16
16
17
17
18
19
19
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1
6.7
6.8
6.9
6.10
xm = b and variants . . . . . . . .
g x = b and variants . . . . . . . .
log(x) = b and variants . . . . . .
Equations involving trigonometric
. . . . . .
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. . . . . .
functions
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20
20
21
22
7 Rewriting expressions
23
8 Derivatives
8.1 Derivatives of power functions . . .
8.2 Derivatives of exponential functions
8.3 The chain rule . . . . . . . . . . . .
8.4 The product rule . . . . . . . . . .
8.5 The quotient rule . . . . . . . . . .
24
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25
25
26
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9 Applications of derivatives
27
9.1 Tangent lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
9.2 Extreme values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
9.3 Inflection points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
A Answers
A.1 Answers
A.2 Answers
A.3 Answers
A.4 Answers
A.5 Answers
A.6 Answers
A.7 Answers
A.8 Answers
A.9 Answers
for
for
for
for
for
for
for
for
for
Section
Section
Section
Section
Section
Section
Section
Section
Section
1
2
3
4
5
6
7
8
9
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2
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30
30
30
31
31
32
32
34
34
36
1
1.1
Expressions with brackets
Expanding brackets
Examples:
x(7 + 5x3 ) = x · 7 + x · 5x3 = 7x + 5x4
(2a − 3)(5b + 2) = 2a · 5b + 2a · 2 + −3 · 5b + −3 · 2 = 10ab + 4a − 15b − 6
(2x + 7)(5x − 11) = 2x · 5x + 2x · −11 + 7 · 5x + 7 · −11 = 10x2 − 22x + 35x − 77 = 10x2 − 13x − 77
Exercises:
Expand brackets in the following expressions. Simplify your result as much as possible.
1. x(2x2 + 7x5 )
6. (x + 6)(x + 7)
11. (x + 3)(2x5 + x2 − 4x)
2. x2 (4x + x2 )
7. (3x + 5)(2x − 7)
12. (2x + 3)(2x3 + 5x2 + 1)
3. 5x(2x2 − 3x + 1)
8. (2x − 5)(5x − 2)
13. (2x2 + 3)(5x2 − 2x + 5)
4. 3x3 (2x2 − 3x + 1)
9. (a + b)(a + b)
14. (2x2 −3x+1)(x5 −2x2 −7x)
10. (a + b)(a − b)
15. (3x2 − 5)(x2 − 4x)(2x + 5)
5. (3x5 + 6x3 − 1) · 4x2
1.2
Special products
Theory:
It is convenient to memorize the following identities and being able to apply them.
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(a + b)(a − b) = a2 − b2
Example:
(3x − 2y)2 = (3x)2 − 2 · 3x · 2y + (2y)2 = 9x2 − 12xy + 4y 2
Exercises:
Expand brackets in the following expressions. Simplify your result as much as possible.
16. (5x + y)2
20. (x + 5)(x − 5)
24. (p + q)2 + (p − q)2
17. (3x − 7y)2
21. (2x − 3)(2x + 3)
25. (p + q)2 − (p − q)2
18. (x2 + 3x)2
22. (x3 + 2x)(x3 − 2x)
26. (p + q + r)2
19. (2x3 − 5x2 )2
23. (x4 + x2 )(x4 − x2 )
27. (x − y)3
1.3
Factorizing
Theory:
In the previous section we have expanded brackets. Sometimes it is necessary to do the opposite,
i.e., to write a sum as a product of two (or more) factors.
3
Example:
In some cases we can take out a common factor:
3x4 − 7x3 + 5x2 = x2 (3x2 − 7x + 5)
In some cases we can use the identity (x + a)(x + b) = x2 + (a + b)x + ab to write the quadratic
expression x2 + px + q in terms of two linear factors: find a and b such that a + b = p and
ab = q.
For example, if we want to factorize x2 + 7x + 10, we can try to find numbers a and b such that
a + b = 7 and ab = 10. In this example we can take a = 5 and b = 2. (Note that a = 2 and
b = 5 also works.) This gives x2 + 7x + 10 = (x + 5)(x + 2).
Exercises:
Factorize the following expressions as much as possible.
28. 4x7 + 3x6 + 7x4
32. x2 + 9x + 20
36. x2 − 12x + 35
29. x5 + 8x2 + 9x
33. x2 + 12x + 20
37. x2 − 7x + 12
30. 2x8 − 5x5 + x3
34. x2 − 4x − 21
38. x2 + 10x + 25
31. −x5 + 4x4 − 7x2 + 9x
35. x2 + 6x − 16
39. x2 − 36
4
2
Fractions
2.1
Computing with fractions
Addition and subtraction:
Two fractions can only be added or subtracted if they have the same denominator. If this is
not the case then we have to multiply the numerator and denominator of each fraction with
the denominator of the other fraction.
Examples:
2
5
11 × 2
7×5
22 25
57
+
=
+
=
+
=
7 11
11 × 7 7 × 11
77 77
77
7 2
5×7 9×2
35 18
17
− =
−
=
−
=
9 5
5×9 9×5
45 45
45
4
2
4
36 4
180 68
112
27
2
−1 =2 − =
− =
−
=
=1
17
5
17 5
17 5
85
85
85
85
1 1
3 1
4
2
+ = + = =
2 6
6 6
6
3
2
2
2
4
4
Remark: in the third example 3
denotes 3 +
and not 3 ∗ . Similarly 1 = 1 + .
17
17
17
5
5
In the Netherlands, it is tradition to use this notation in case of fractions with numbers only.
3x
x
.
In other expressions generally multiplication is meant, e.g. 3x = 3 ∗ x and 3 means
y
y
3
Exercises:
Write the following expressions as a single fraction. Simplify the result as much as possible.
1.
5 2
+
8 7
5.
5 2
−
8 7
2
1
9. 4 + 2
7
5
2.
3
3
+
7 11
6.
3
3
−
7 11
10. 8
3
1
−5
11
3
3.
5
7
+
18 18
7.
5
7
−
18 18
11. 1
1
7
+5
18
6
4.
8 7
+
9 8
8.
8 7
−
9 8
1
3
12. 5 − 1
9
8
Multiplication and division:
The product of two fractions can be computed by just multiplying numerators and denominators. Dividing by a fraction p/q is the same as multiplying by its reciprocal q/p.
Examples:
3
7
3×7
21
× =
=
19 5
19 × 5
95
3 7
3
5
15
: =
× =
19 5
19 7
133
1
1
7 5
35
11
2 ×1 = × =
=2
3
4
3 4
12
12
5
Exercises:
Write the following expressions as a single fraction. Simplify the result as much as possible.
13.
5 2
×
8 7
17.
5 2
:
8 7
1
21. 2 : 6
2
2
1
25. 4 × 2
7
5
14.
3
3
×
7 11
18.
3 3
:
7 11
3
22. 1 : 2
7
26. 8
3
1
×1
11
3
15.
7 5
×
8 6
19.
7 5
:
8 6
23.
7 8
×
8 7
27. 1
1
7
:1
18
6
16.
8 7
×
9 8
20.
8 7
:
9 8
8
24. 3 × 3
9
2.2
1
1
28. 5 : 1
9
3
Fractions with variables
Theory:
Fractions containing variables can be treated in the same way as fractions consisting of only
numbers.
Examples:
b
11 × a
7×b
11a 7b
11a + 7b
a
+
=
+
=
+
=
7 11
11 × 7 7 × 11
77
77
77
2c 7
2c × 7
14c
× =
=
3
b
3×b
3b
3d 11
3d
d
3d2
:
=
×
=
5
d
5
11
55
4
5(x−3)
4(x+1)
5x − 15
4x + 4
9x − 11
5
+
=
+
=
+
=
x+1 x−3
(x+1)(x−3) (x+1)(x−3)
(x+1)(x−3) (x+1)(x−3)
(x+1)(x−3)
Exercises:
Write the following expressions as one fraction. Simplify the result as much as possible.
29.
p 2q
+
8
7
33.
5a + 3 2
×
8
7
37.
x+1 a
+
7
5
30.
3t
3
+
7
11
34.
3p 3q
×
7
8
38.
3
1
+
p + 3 3p
31.
7
3
+
x−1 x+6
35.
3p
3q
:
7 q+1
39.
7
11
−
3q + 1 q + 3
32.
y y
−
8 9
36.
x
2
:
3 x+3
40.
a
2a − 1
−
2x + 3 4x + 1
2.3
Solving equations with fractions
Theory:
Equations involving fractions can be solved by cross multiplication:
A
P
=
B
Q
⇔
A×Q=B×P
(provided that both B 6= 0 and Q 6= 0).
6
Caution: after solving the equation we always have to check wether the conditions B 6= 0 and
Q 6= 0 are satisfied!
Example:
Using cross multiplication we can solve the following equation:
x−2
x+2
=
⇔
3
5
⇔
⇔
⇔
(x − 2) · 5 = 3 · (x + 2)
5x − 10 = 3x + 6
2x = 16
x=8
In this example it is obvious that we do not have to check whether the denominator is zero for
some x.
Exercises:
Solve the following equations.
41.
x+7
x+2
=
3
2
45.
x x−5
x
+
=
4
7
3
42.
x+3
x−1
=
2
4
46.
x+1 x+9
7x + 3
+
=
2
3
4
43.
6
9
=
x+1
x
47.
x+5 x
− =x−5
2
3
44.
3x − 1
x+1
=
5
2
48.
x+7
x+1
=
x−1
x−3
7
3
Square roots
Theory:
√
√ √
For nonnegative numbers a and b we have ab = a b. This rule can be used to rewrite
expressions involving square roots.
Examples:
√
√
√
√
√
75 = 25 · 3 = 25 × 3 = 5 3
√
√
√
√
√ √
√ √
√
√
√
27 − 12 = 9 · 3 − 4 · 3 = 9 3 − 4 3 = 3 3 − 2 3 = 3
Exercises:
Rewrite the following expressions as in the preceding examples.
√
√ √
1. 32
9. 3 · 12
√
√ √
2. 175
10. 6 · 15
√
√
√
3. 176
11. 10 · 98
√
√
√
4. 147
12. 30 · 105
√
√
√
√
5. 18 + 50
13. (3 + 7)(3 − 7)
√
√
√
6. 20 + 180
14. (2 + 2)2
√
√
√
√
7. 363 − 192
15. ( 7 − 5)2
√
√ √
√
√
√
8. 128 − 50
16. ( 12 − 6)( 3 + 6)
8
4
4.1
Exponents and logarithms
Computing with exponents
Theory:
If x, y, g > 0 then the following rules apply:
xa · y a
xa
ya
ga · gb
ga
gb
(g a )b
1
gk
= (x · y)a
a
x
=
y
a+b
= g
= g a−b
= g a·b
= g −k
Examples:
73 × 75 = 73+5 = 78
37 : 39 = 37−9 = 3−2 (= 1/9)
(a5 )3 = a5×3 = a15
Exercises:
Rewrite the following expressions as a single power.
1. 58 × 53
5. a3 · a2
9. (35 )4
2. 62 × 67
6. b4 · b−7
10. (x2 )7
14.
3. 37 : 32
7. c5 : c3
11. (y 3 · y 2 )3
15. 55 : 25
4. 77 : 79
8. d3 : d7
12. (z 5 ·
4.2
13. 25 × 82
1 7
)
z3
1
7
× 493
16. (x3 )2 : x6
Roots as exponents
Theory:
√
√
1
1
For x ≥ 0 we have n x = x n . In particular we have x = x 2 .
Examples:
√
1
5
9 3 = 32 · 3 2 = 3 2
5
x3
x3
x3
3− 34
3
√
=
=
=
x
=
x
1
4
x3x
x · x3
x3
Exercises:
Rewrite the following expressions as a single power. (Assume that all variables that appear are
strictly positive.)
9
√
17. 125 · 5
√
18. 27 · 3 9
√
19. 16 : 2
√
7
20.
49
4.3
√
3
p
√
3
x2 · x
√
26. (x · 3 x)4
√
√
3
27. x2 : x3
74
√
5
22. a a2
√
3
23. b3 b5
21.
24. c ·
25.
√ √
c: 3c
28.
x2
1
√
· x
Solving exponential equations
Theory:
If g > 0 and g 6= 1 then g A = g B implies that A = B. We can use this rule to solve some
exponential equations analytically. Caution: in order to solve an exponential equation we must
first write both sides of the equation in terms of the same base number!
Example:
To solve the following equation we write each side as a power of 2:
√
1
8x = 4 2 ⇔ (23 )x = 22 · 2 2
5
⇔ 23x = 2 2
5
⇔ 3x =
2
5
⇔ x=
6
Exercises:
Solve the following equations.
30. 3x+1 = 27
√
33. 4a+1 = 2 2
√
34. 7 × 49b = 7 3 7
31. 52x = 125
35. 125 × 25c+3 =
32. 73x+1 = 49
36. 2434d+1 = 27
29. 2x = 512
4.4
37. 23x−2 = 4x
1
5
38. 9 × 3y+5 = 27y−1
√
39. 5z+7 = 125 × 5z
40. (3 · 9x−1 )3 =
Logarithms: applying the definition
Theory:
The logarithm with base number g is defined by the equivalence
ga = b
⇔
logg (b) = a
In particular, this implies that
logg (g a ) = a and g logg (b) = b
Examples:
log3 (81) = log3 (34 ) = 4
√
5
1
5
log5 (25 5) = log5 (52 · 5 2 ) = log5 (5 2 ) =
2
Exercises:
Compute the logarithms and solve the equations.
10
1
√
3
3
41. log2 (32)
45. log4 (2)
49. log5 (a) = 2
42. log3 (243)
√
43. log7 (49 7)
46. log9 ( 13 )
50. log3 (b) = 4
47. log8 (32)
51. log9 (c) =
1
)
44. log5 ( 25
48. log2 (8 5 )
4.5
7
5
2
52. log6 (d) = −2
Logarithms: applying the rules
Theory:
Logarithms are just exponents. Hence, the rules for computing with logarithms correspond to
those for exponents:
logg (A) + logg (B) = logg (A · B)
A
logg (A) − logg (B) = logg ( )
B
n · logg (A) = logg (An )
logp (A)
logg (A) =
logp (g)
Examples:
log3 (x) + log3 (x + 3) = log3 (x · (x + 3)) = log3 (x2 + 3x)
log2 (5x) + 3 · log2 (x) = log2 (5x) + log2 (x3 ) = log2 (5x · x3 ) = log2 (5x4 )
3 + log2 (x − 3) = log2 (23 ) + log2 (x − 3) = log2 (8) + log2 (x − 3) = log2 (8(x − 3)) = log2 (8x − 24)
Exercises:
Rewrite the following expressions as a single logarithm.
53. log3 (x + 5) + log3 (x − 1)
57. 4 + log3 (2x − 5)
54. 3 · log2 (x) + log2 (x + 5)
58.
55. log5 (2x + 7) − log5 (7x + 2)
59. log6 (3x + 2) − 1
56. 2 · log5 (x + 3) − log5 (x − 9)
60. log9 (x − 3) + log3 (x + 3)
11
1
2
+ log4 (3x + 5)
5
5.1
Trigonometric functions
Relations between sine and cosine
Theory:
The sine and cosine functions are periodic, with period 2π. This means that every 2π the
function repeats itself (see the figures below). The cosine function can be obtained from the
sine function by shifting the graph of the sine function 12 π to the left. Similarly, the sine function
can be obtained by shifting the graph of the sine function 12 π to the right. In formulae:
sin(x + k∗2π) = sin(x),
cos(x + k∗2π) = cos(x),
for any integer k (k = · · · − 2, −1, 0, 1, 2, · · · )
for any integer k (k = · · · − 2, −1, 0, 1, 2, · · · )
1
cos(x) = sin(x + π)
2
1
sin(x) = cos(x − π)
2
Figure 1: The sine and cosine function are periodic, with period 2π. There is a horizontal shift
of 12 π between the two curves.
12
Exercises:
Complete the following identities. Each time use one of the following options:
sin(φ), − sin(φ), cos(φ), or − cos(φ).
1. sin(−φ) =
5. sin( 12 π + φ) =
2. sin(π − φ) =
6. cos(−φ) =
3. sin(π + φ) =
7. cos(π − φ) =
4. sin( 12 π − φ) =
8. cos(π + φ) =
5.2
9. cos( 21 π − φ) =
10. cos( 12 π + φ) =
Exact values
Theory:
For a number of special angles, the sine, cosine and tangent function have ’nice’ values. It is
convenient to memorize the following values:
0
1
π
6
1
π
4
1
π
3
1
π
2
sin
0
1
2
√
1
2
2
√
1
3
2
1
cos
1
√
1
3
2
√
1
2
2
1
2
0
tan
0
√
1
3
3
1
√
3
not defined
√
If you remember the special angles, then fill in the column for sine by means of 21 0 = 0,
√
√
√
1
1 = 21 , 21 2, 12 3. The colum for the cosine is the reverse. The column for the tangent can
2
be obtained by using
sin(x)
tan(x) =
cos(x)
Figure 2: The tangent function is periodic, with period π, i.e. tan(x + kπ) = tan(x).
13
Exercises:
Give exact values for the following expressions.
11. sin( 13 π)
15. cos( 13 π)
19. tan( 13 π)
12. sin( 56 π)
16. cos( 56 π)
20. tan( 56 π)
13. sin(3π)
17. cos(3π)
21. tan(3π)
14. sin(− 43 π)
18. cos(− 34 π)
22. tan(− 43 π)
5.3
Inverse trigonometric functions
Theory:
The inverse trigonometric functions arcsin, arccos and arctan function are the reverse of the
sin, cos and tan functions, respectively. Since the trigonometric functions are periodic, the
inverse trigonometric functions would have more than one function value assigned to one x
value. Therefore, the inverse trigonometric functions are usually restricted to their principal
values: − 21 π ≤ arcsin(x) ≤ 12 π , 0 ≤ arccos(x) ≤ π , − 12 π ≤ arctan(x) ≤ 12 π ,
Example:
To determine the value of arccos( 21 ) we ask ourselves for which α, cos(α) = 12 , with 0 ≤ α ≤ π.
Since cos( 13 π) = 12 the outcome is arccos( 21 ) = 13 π.
Exercises:
Give exact values for the following expressions.
23. arcsin( 21 )
25. arccos(0)
24. arcsin(−1)
26. arccos(− 2 2)
5.4
√
1
27. arctan(1)
√
28. arctan( 3)
Addition formulas
Theory:
It is convenient to memorize the following identities:
cos(α + β)
sin(α + β)
2
sin (α) + cos2 (α)
sin(2α)
cos(2α)
=
=
=
=
=
cos(α) cos(β) − sin(α) sin(β)
sin(α) cos(β) + cos(α) sin(β)
1
2 sin(α) cos(α)
2 cos2 (α) − 1 = 1 − 2 sin2 (α) = cos2 (α) − sin2 (α)
Example:
We can use trigonometric identities to compute exact values of sine and cosine for non-standard
angles. For example, we have
5
cos( 12
π) = cos( 16 π + 14 π)
= cos( 61 π) cos( 14 π) − sin( 61 π) sin( 14 π)
√
√
√
= 12 3 · 21 2 − 12 · 12 2
√ √
= 14 2( 3 − 1)
14
Exercises:
Use trigonometric identities to compute exact values for the following expressions.
7
π)
29. sin( 12
1
30. cos( 12
π)
31. sin( 18 π)
15
6
6.1
Solving equations
Linear equations
Exercises:
Solve the following equations.
9. 7h − 8 = 4h + 7
1. 5a + 7 = 22
5. 4(e + 2) = 28
2. 3x − 8 = 25
6. 5(3f + 11) = 10
10. 4(k + 3) = k + 6
3. 19 − 7c = 5
7. 3(2g − 3) = 5
11. 7(3m + 2) = 2m + 5
4. 16 + 9d = 7
8. 14(g + 3) = 40
12. 20(n − 1) = 9(n − 2)
6.2
Quadratic equations
Quadratic equations are equations of the form ax2 + bx + c = 0 with a 6= 0. We can either try
to factorize this equation or we can use the quadratic formula (also known as the abc-formula).
6.2.1
Factorizing
Theory:
If a quadratic expression can be factorized in terms of two linear factors then we can use the
rule
A · B = 0 ⇔ A = 0 or B = 0
which reduces the quadratic equation to solving two linear equations.
Examples:
To solve the equation x2 − 3x − 10 = 0 we can try to find two numbers a and b such that
a + b = −3 and ab = −10. For example, we can take a = −5 and b = 2. Then
x2 − 3x − 10 = 0 ⇔ (x − 5)(x + 2) = 0
⇔ x − 5 = 0 or x + 2 = 0
⇔ x = 5 or x = −2
Here is a more complicated example:
3x2 − 26 = (x − 2)2 ⇔
⇔
⇔
⇔
⇔
⇔
Exercises:
Solve the following equations.
16
3x2 − 26 = x2 − 4x + 4
2x2 + 4x − 30 = 0
x2 + 2x − 15 = 0
(x + 5)(x − 3) = 0
x + 5 = 0 or x − 3 = 0
x = −5 or x = 3
13. (x − 7)(x − 3) = 0
17. x2 = 2x + 3
21. 5x2 − 7x + 3 = (2x − 1)2
14. x2 + 7x + 10 = 0
18. x2 + 3x = 7(x + 11)
22. 3x2 + 8x + 5 = x2 + 4x + 11
15. x2 + 4x − 5 = 0
19. x2 + 15x + 6 = 8x − 4
23. 3x2 + 6x + 1 = (x + 3)2
16. x2 − 9x + 18 = 0
20. x2 − 3x + 19 = 5x + 3
24. (2x + 3)2 = (x − 6)2
6.2.2
The quadratic formula (also known as the abc-formula)
Theory:
If the method of factorizing does not work we can also use the following formula:
√
−b ± b2 − 4ac
2
ax + bx + c = 0 ⇔ x =
2a
Examples:
For the equation 2x2 − 7x − 15 = 0 we have a = 2, b = −7, and c = −15. Substituting these
values into the quadratic formula gives
√
√
√
7 ± 49 + 120
7 ± 169
7 ± 13
7 ± 49 − 4 · 2 · −15
=
=
=
.
x=
2·2
4
4
4
This gives the solutions
x=
7 − 13
3
=−
4
2
and x =
7 + 13
= 5.
4
For the equation 3x2 + 5x + 1 = 0 we have a = 3, b = 5, and c = 15. Substituting these values
into the quadratic formula gives
√
√
√
−5 ± 25 − 12
−5 ± 13
−5 ± 25 − 4 · 3 · 1
=
=
.
x=
6
6
6
This gives the solutions
√
−5 − 13
x=
6
√
−5 + 13
and x =
.
6
Exercises:
Solve the following equations.
25. 3x2 + 7x − 6 = 0
29. 3x2 + 8x + 2 = 0
26. 4x2 + 16x + 15 = 0
30. 2x2 − 5x − 4 = 0
27. 4x2 + x − 3 = 0
31. 5x2 + x − 1 = 0
28. 5x2 + 21x + 4 = 0
32. 3x2 + 3x + 2 = 0
6.3
Equations of higher degree
Theory:
The main idea is to write a sum of several terms as a product of two (or more) factors. The
17
product equals zero if and only if one of the factors equals zero. Hence, factorization can be
used to reduce an equation into two or more equations having a lower degree.
Examples:
The equation x6 + 5x5 + 6x4 = 0 can be solved by taking out the common factor x4 :
x6 + 5x5 + 6x4 = 0 ⇔
⇔
⇔
⇔
x4 (x2 + 5x + 6) = 0
x4 (x + 3)(x + 2) = 0
x4 = 0 or x + 3 = 0 or x + 2 = 0
x = 0 or x = −3 or x = −2
The equation x6 + 7x3 − 8 = 0 can be considered as a quadratic equation in x3 . Rewrite the
equation by setting y = x3 :
x6 + 7x3 − 8 = 0 ⇔
⇔
⇔
⇔
⇔
⇔
y 2 + 7y − 8 = 0
(y + 8)(y − 1) = 0
y + 8 = 0 or y − 1 = 0
y = −8 or y = 1
x3 = −8 or x3 = 1
x = −2 or x = 1
Exercises:
Solve the following equations.
33. x5 − 7x4 + 12x3 = 0
37. x5 − 5x3 + 4x = 0
34. 7x3 + 5x2 = 0
38. x4 − 8x2 = 9
35. x3 = 4x
39. 4x3 + 11x2 = 3x
36. x4 − 5x2 + 4 = 0
40. 4x4 + 11x2 = 3
6.4
A2 = B 2
Theory:
The equality A2 = B 2 implies A2 − B 2 = 0 or, equivalently, (A + B)(A − B) = 0. This implies
that either A = −B or A = B.
Example:
Applying the previous rule gives:
(x2 + 3x)2 = (3x + 8)2 ⇔
⇔
⇔
⇔
x2 + 3x = 3x + 8 or x2 + 3x = −3x − 8
x2 = 8 or x2 + 6x + 8 = 0
x2 = 8 or (x + 4)(x + 2) = 0
√
√
√
√
x2 = − 8 = −2 2 or x = 8 = 2 2 or x = −4 or x = −2
Exercises:
Solve the following equations.
41. (2x + 5)2 = (x + 1)2
18
42. (7x − 2)2 = (3x − 10)2
43. (2x2 + x − 31)2 = (x2 − x + 4)2
44. (3x2 + 2x + 1)2 = (2x2 + 5x − 1)2
6.5
A·B =A·C
Theory:
The equation A · B = A · C implies A · B − A · C = 0 or, equivalently, A · (B − C) = 0. This
implies that either A = 0 or B = C.
Example:
(x3 − 8)(3x − 5) = (x3 − 8)(x + 1) ⇔ x3 − 8 = 0 or 3x − 5 = x + 1
⇔ x3 = 8 or 2x = 6
⇔ x = 2 or x = 3
Exercises:
Solve the following equations.
45. (x − 5)(x5 + 7) = (x − 5)(5x2 + 7)
46. (x3 + 1)(x3 − 2x2 ) = (x3 + 1)(2x2 + 5x)
47. (7x + 5)(2x + 8) = (2x + 7)(2x + 8)
48. (x2 − 9)3 (x − 3) = (x2 − 9)(x + 7)
6.6
√
A=B
Theory:
√
Equations of the form A = B can be solved by squaring both sides:
√
A = B ⇒ A = B2.
Caution: if B < 0 the reverse implication does not hold! Therefore, we must always check the
final solutions whenever we square both sides of an equation!
Example:
To solve the following equations we first square both sides:
√
2x − 1 = 2x − 7 ⇒ 2x − 1 = (2x − 7)2
⇒ 2x − 1 = 4x2 − 28x + 49
⇒ 4x2 − 30x + 50 = 0
⇒ 2x2 − 15x + 25 = 0
⇒ (2x − 5)(x − 5) = 0
5
⇒ x = or x = 5
2
19
Substituting x = 25 in the original equation shows that the right hand side is negative. Hence,
x = 5 is the only solution.
Exercises:
Solve the following equations.
√
49. 5x + 1 = x − 1
√
50. x2 + 7 = x + 1
√
51. 2x2 + 4 = x + 2
√
52. 2x2 + 4 = 2x − 2
√
53. x + 4x − 3 = 2
√
54. 8x + 1 = x + 2
√
55. 3 + 2x2 − 1 = 2x
√
56. 11x + 5 = x + 3
xm = b and variants
6.7
Theory:
Using the rules for exponents, the equation xm = b can be reduced to
(xm )1/m = b1/m
⇒
x = b1/m .
Caution: if m is an even integer, then the graph of y = xm is symmetric around the y-axis and
lies above the x-axis. In this case, the equation xm = b has two solutions if b > 0 (x = b1/m
and x = −b1/m ) and no solutions if b < 0.
Example:
√
2
1
5
Solving x 2 = 39 gives x = 39 5 = (392 ) 5 = 5 1521.
Exercises:
Solve the following equations.
57. a3 = 17
58. b6 = 49
3
59. c 2 = 4
60. d−3 = 11
6.8
g x = b and variants
Theory:
Mind the difference with the preceding section: now the unknown is the exponent instead of
the base number!
In Section 4.3 this type of equation has been solved by writing b as a power of the base number
g. If that does not work, then we have to use logarithms: g a = b ⇔ logg (b) = a.
Example: Using logarithms we can solve the following equation:
3 · 7x+2 = 39 ⇔ 7x+2 = 13
⇔ x + 2 = log7 (13)
⇔ x = −2 + log7 (13)
20
Exercises:
Solve the following equations.
61. 5x−3 = 123
62. 7 · 22x−6 = 35
63. 8 + 33x+1 = 20
64. 5x + 53 = 55
6.9
log(x) = b and variants
Theory:
Equations involving exponents and logarithms can be solved by using the following rule:
ga = b
⇔
logg (b) = a.
By using the rules for logarithms it is often possible to rewrite a complicated expression in
terms of a single logarithm.
Example:
Using the properties of logarithms gives:
log3 (2x − 1) + log3 (x + 4) = 4 ⇒
⇒
⇔
⇔
⇔
⇔
log3 ((2x − 1)(x + 4)) = 4
(2x − 1)(x + 4) = 34
2x2 + 7x − 4 = 81
2x2 + 7x − 85 = 0
(2x + 17)(x − 5) = 0
x = −8 21 or x = 5
Caution: in the process of solving this equation we have used a rule which only applies for
A, B > 0. Therefore, we have to check the final answers! For x = −8 12 we have 2x − 1 < 0.
Hence, the only solution is x = 5.
Exercises:
Solve the following equations.
65. log2 (3x + 1) + log2 (x − 1) = 6
66. log5 (8x − 3) + log5 (4x − 9) = 3
67. 2 log2 (x) + log2 (x2 − 8) = 7
68. log6 (5x + 1) = 1 + log6 (x − 1)
21
6.10
Equations involving trigonometric functions
Theory:
Since the trigonometric functions are periodic, equations involving trigonometric functions generally lead to periodic solutions.
Example:
Solve cos(2x + π4 ) = 12 .
On the standard interval 0 ≤ α ≤ 2π, the function cos(α) = 21 when α = 13 π or α = 35 π. The
cosine function has period 2π. Therefore, the equation cos(α) = 12 has the (periodic) solutions
α = 13 π + k 2π or α = 35 π + k 2π, where k can be any integer number (positive or negative):
k = · · · − 3, −2, −1, 0, 1, 2, 3, · · · . We can now solve the original problem:
1
π
5
π
= π + k 2π or 2x + = π + k 2π
4
3
4
3
1
17
⇔ 2x = π + k 2π or 2x = π + k 2π
12
12
1
17
⇔ x = π + kπ or x = π + kπ
24
24
2x +
Exercises:
Solve the following equations, for every possible x.
69. sin(x) =
1
2
70. sin(2x) =
1
2
√
3
√
71. cos(x) = − 21 2
72. cos(3x − π6 ) =
1
2
73. tan(x) = 1
74. tan(4x) =
75. sin2 (x) =
√
1
2
3
(i) by using square root
(ii) by using double angle
22
7
Rewriting expressions
Theory:
Formulas describe relationships between different variables. In order to see how one variable
depends on the other variables, it is necessary to isolate this variable from the formula.
Example:
Consider the formula
z
.
x+3
y =p+
Write x as a function of the variables p and z.
y =p+
z
z
⇔ y−p=
x+3
x+3
⇔ (y − p)(x + 3) = z
z
⇔ x+3=
y−p
z
⇔ x=
−3
y−p
Exercises:
1. Isolate p from q =
√
2p + 1
5. Isolate v from E = m · g · h + 12 m · v 2
2. Isolate A from A · B 2 · C 3 = 1
6. Isolate P from I = log2 (P )
3. Isolate R from Q = I 2 · R · t
7. Isolate x from y =
4. Isolate m from E = m · g · h + 21 m · v 2
x
x+2
√
8. Isolate t from y = t + t2 + 1
23
8
Derivatives
Theory:
The derivative f 0 of a function f gives the slope of the graph at each point: f 0 (p) is the slope
of the line tangent to the graph of f at the point (p, f (p)).
8.1
Derivatives of power functions
Theory:
For every exponent n we have:
f (x) = axn
f 0 (x) = naxn−1
⇒
In addition:
f 0 (x) = p0 (x) + q 0 (x)
⇒
f (x) = p(x) + q(x)
Examples:
f (x) = 4x3 + 5x2 − 9x + 3 ⇒ f 0 (x) = 12x2 + 10x − 9
4
8
f (x) = 2 = 4x−2 ⇒ f 0 (x) = −8x−3 = − 3
x
x
Exercises:
Differentiate the following functions.
4
3
+ 3
x x
√
√
8
6. f (x) = 4x2 x + 6 x + √
x
1. f (x) = 6x3 + 3x2 + 4x + 2
5. f (x) =
2. f (x) = 4x5 − 5x3 − x2 + 7x
3. f (x) = x3 · (3x2 + 4x)
7. f (x) = (x2 +
4. f (x) = (2x2 − 3)2
8. f (x) = (x2 + 2x)(x2 − 3x)
8.2
Derivatives of exponential functions
Theory:
For the derivative of an exponential function we have the rule:
f (x) = g x
⇒
f 0 (x) = g x · ln(g)
In particular, we have:
f (x) = ex
⇒
Exercises:
Differentiate the following functions.
9. f (x) = 4ex + 8
10. f (x) = 3x + 5x
11. f (x) = 2−x
12. f (x) = 3 · ex + e · 3x + x · 3e
24
f 0 (x) = ex
1 2
)
x2
8.3
The chain rule
Theory:
If we compute the value of a function f (x) by first computing h(x) and then substituting the
result into another function k(x), then f is the so-called composition of k and h:
f (x) = k(h(x))
De derivative of f is computed by means of the chain rule:
f 0 (x) = k 0 (h(x))h0 (x)
Example:
The function f (x) = 3(x3 − 2x + 5)4 is the composition of the functions h(x) = x3 − 2x + 5 and
k(u) = 3u4 . Hence, we obtain
h0 (x) = 3x2 − 2
k 0 (u) = 12u3
f 0 (x) = k 0 (h(x))h0 (x)
= 12(h(x))3 · h0 (x)
= 12(x3 − 2x + 5)3 · (3x2 − 2)
Exercises:
Differentiate the following functions.
13. f (x) = 5(x2 − 2x)7
1
+ 5x − 1
√
15. f (x) = 6 x2 + 1
14. f (x) =
x3
x3 +x
16. f (x) = e
8.4
1
17. f (x) = 2
(x + 3x + 1)3
18. f (x) = √
1
√
21. f (x) = e x
√
22. f (x) = ex
+1
23. f (x) =
19. f (x) = e1/x
√
20. f (x) = x + ex
1
e2x + x2
24. f (x) =
1
√
x3 + e x + x
x2
The product rule
Theory:
Products of functions are differentiated by means of the product rule:
f (x) = p(x) · q(x)
⇒
f 0 (x) = p0 (x)q(x) + p(x)q 0 (x)
Example:
The function f (x) = x3 ·(x2 +x)4 is the product of p(x) = x3 and q(x) = (x2 +x)4 . Differentiating
the factors gives
p0 (x) = 3x2 and q 0 (x) = 4(x2 + x)3 · (2x + 1)
Hence, applying the product rule gives
f 0 (x) = 3x2 · (x2 + x)4 + x3 · 4(x2 + x)3 · (2x + 1)
With some additional manipulations this expression can be simplified.
Exercises:
Differentiate the following functions.
25
25. f (x) = (3x3 + 4x2 + 1)(2x4 + 3x + 5)
29. f (x) = xex
26. f (x) = x(x + 5)7
30. f (x) = ex (x3 + x2 + x + 1)
√
31. f (x) = x2 x + ex
√
32. f (x) = ex x2 + 2x
27. f (x) = x3 (x3 − 3x2 + 6)5
√
28. f (x) = x5 4x − 1
8.5
The quotient rule
Theory:
Quotients of functions are differentiated by means of the quotient rule:
f (x) =
p(x)
q(x)
⇒
f 0 (x) =
p0 (x)q(x) − p(x)q 0 (x)
q(x)2
Example:
The function
3x2
x3 + x + 1
is the quotient of p(x) = 3x2 and q(x) = x3 + x + 1. Hence, the quotient rule gives
f (x) =
f 0 (x) =
6x(x3 + x + 1) − 3x2 (3x2 + 1)
(x3 + x + 1)2
With some additional manipulations this expression can be simplified further.
Exercises:
Differentiate the following functions.
x2
33. f (x) =
2x + 1
34. f (x) =
x2 + 1
x2 − 1
2x3 − 3x2
35. f (x) =
x+3
36. f (x) =
x3 − 3
x2 − 2
ex
37. f (x) = x
e +1
38. f (x) =
x3 (2x + 1)2
x3 − 1
(2x + 5)3
39. f (x) =
ex
√
x+ x
√
40. f (x) =
1+ x
26
9
9.1
Applications of derivatives
Tangent lines
Theory:
The value of f 0 (p) equals the slope of the tangent line to the graph of f at the point P (p, f (p)).
By substituting the coordinates of the point P in the equation y = f 0 (p) · x + b we can compute
the value of b. In this manner we can find the equation of the tangent line.
Example:
Consider the function f (x) = x2 − 9/x and the point P having x-coordinate 3. The slope of the
tangent line equals f 0 (3). The derivative of f is given by f 0 (x) = 2x + 9/x2 , so that f 0 (3) = 7.
In addtion, we have f (3) = 6 so the y-coordinate of P equals 6. Plugging the coordinates into
the equation y = 7x + b gives 6 = 7 × 3 + b, which gives b = −15. Hence, the equation of the
tangent line is y = 7x − 15.
Exercises:
Determine the tangent line to the given functions in the indicated points.
√
5. f (x) = x2 − 4 x ; P (1, −3)
1. f (x) = x3 − 4x2 + x − 2 ; P (4, 2)
2. f (x) = x2 + 3x −
8
x
√
6. f (x) = x2 x − 1 ; P (2, 4)
; P (2, 6)
4x2 − 6
; P (3, 3)
x2 + 1
√
4. f (x) = 7 · 4x2 − 10x − 1 ; P (5, 49)
3. f (x) =
9.2
7. f (x) = x2 ex
; P (1, e)
8. f (x) = x(2x − 5)3
; P (0, 0)
Extreme values
Theory:
If f is a function, then f (p) is called a (local) maximum if f (x) ≤ f (p) for all values x close
to p. The value f (p) is called a (local) minimum if f (p) ≤ f (x) for all values x close to p.
Maxima and minima are called extreme values.
We can find extreme values for f by solving the equation f 0 (x) = 0. Caution: satisfying the
equation f 0 (x) = 0 does not guarantee that f (x) is an extreme value. Conversely, a function f
can have an extreme value f (x), but with f 0 (x) 6= 0!
Example:
√
Consider the function f (x) = x2 − 2x x − 2x. We now want to find the extreme values of f .
Note that f is only defined for x ≥ 0.
√
0
0
The derivative
√ is given by f (x) = 2x − 3 x − 2. The equation f (x) = 0 is easier to solve by
writing p = x:
f 0 (x) = 0
⇔
⇔
⇔
⇔
2p2 − 3p − 2 = 0
(p + 12 )(p − 2) = 0
p = − 12 or p = 2
√
√
x = − 12 or x = 2
27
√
Only the second equation gives a solution: from x = 2 it follows that x = 4. Hence, in the
point T (4, f (4)) = (4, −8) the graph of f has a horizontal tangent line. Now compute several
values of f to make a rough sketch of its graph:
f (0) = 0,
f (4) = −8,
f (9) = 9.
A sketch of the graph of f then shows that f (4) = −8 is a minimum. In addtion, we have
f 0 (0) = −2 < 0. So the graph of f decreases in the point (0, 0). This means that f (0) = 0 is
the largest value of f (x) for points x close to 0. Hence, f (0) = 0 is a maximum. (Note that
f 0 (0) 6= 0!)
Exercises:
Compute the extreme values of the following functions. Also check whether it is a maximum
or a minimum.
9. f (x) = x2 − 8x + 11
10. f (x) = 31 x3 − 4x2 + 12x
√
11. f (x) = x2 x + 5
12. f (x) =
9.3
1
1
−
x + 3 4x
Inflection points
Figure 3: The function f (x) = x3 is concave downward for x < 0 and concave upward for
x > 0. The point x = 0 is an inflection point.
Theory:
The sign of a function f and its first and second derivative provide a lot of information about
the shape of its graph:
• The sign of f indicates whether its graph is above (f (x) > 0) or below (f (x) < 0) the
x-axis.
28
• The sign of f 0 indicates whether f increases (f 0 (x) > 0) or decreases (f 0 (x) < 0).
• The sign of f 00 indicates whether tangent lines are above or below the graph of f :
– If f 00 (x) > 0 then f 0 is increasing and tangent lines lie below the graph of f . In this
case the graph of f is concave upward.
– If f 00 (x) < 0 then f 0 is decreasing and tangent lines lie above the graph of f . In this
case the graph of f is concave downward.
– A point where the graph switches from concave downward to upward (or vice versa)
is called an inflection point (in Dutch: buigpunt). These points can be found by
solving the equation f 00 (x) = 0.
For example, the function f (x) = x3 has an inflection point at x = 0, see Figure 3.
Example:
Consider the function f (x) = 13 x3 + 2x2 − 5x + 1. The first and second derivative are given by
f 0 (x) = x2 + 4x − 5 and f 00 (x) = 2x + 4.
Solving f 00 (x) = 0 gives x = −2. Hence, f has an inflection point at (−2, f (−2)) = (−2, 15 31 ).
Exercises:
Compute for the following functions the inflection points of their graph.
13. f (x) = x4 − 6x2 + x − 7
14. f (x) = (x2 − 6x + 7)ex
√
15. f (x) = (x + 6) x
16. f (x) =
x
x2 + 3
29
A
Answers
A.1
Answers for Section 1
1. 2x3 + 7x6
21. 4x2 − 9
2. 4x3 + x4
22. x6 − 4x2
3. 10x3 − 15x2 + 5x
23. x8 − x4
4. 6x5 − 9x4 + 3x3
24. 2p2 + 2q 2
5. 12x7 + 24x5 − 4x2
25. 4pq
2
6. x + 13x + 42
26. p2 + q 2 + r2 + 2pq + 2pr + 2qr
7. 6x2 − 11x − 35
27. x3 − 3x2 y + 3xy 2 − y 3
8. 10x2 − 29x + 10
28. x4 (4x3 + 3x2 + 7)
9. a2 + 2ab + b2
2
10. a − b
29. x(x4 + 8x + 9)
2
6
5
3
30. x3 (2x5 − 5x2 + 1)
2
11. 2x + 6x + x − x − 12x
4
3
31. x(−x4 + 4x3 − 7x + 9)
2
12. 4x + 16x + 15x + 2x + 3
32. (x + 4)(x + 5)
13. 10x4 − 4x3 + 25x2 − 6x + 15
14. 2x7 − 3x6 + x5 − 4x4 − 8x3 + 19x2 − 7x
33. (x + 2)(x + 10)
15. 6x5 − 9x4 − 70x3 + 15x2 + 100x
34. (x + 3)(x − 7)
16. 25x2 + 10xy + y 2
35. (x − 2)(x + 8)
17. 9x2 − 42xy + 49y 2
36. (x − 7)(x − 5)
18. x4 + 6x3 + 9x2
37. (x − 3)(x − 4)
19. 4x6 − 20x5 + 25x4
38. (x + 5)2
20. x2 − 25
39. (x − 6)(x + 6)
A.2
Answers for Section 2
1.
51
56
5.
19
56
9. 6
19
35
2.
54
77
6.
12
77
10. 2
31
33
3.
2
3
7.
1
9
11. 6
5
9
8.
1
72
12. 3
53
72
4. 1
55
72
30
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
5
28
9
77
35
48
7
9
3
2
16
4
1
7
1
1
20
1
1
63
5
12
5
7
1
2
11
3
A.3
25. 9
26. 11
1
33
36.
x2 + 3x
6
37.
5x + 7a + 5
35
27. 1
4
21
38.
10p + 3
3p(p + 3)
28. 3
5
6
39.
−26q + 10
(3q + 1)(q + 3)
2x − 5a + 3
(2x + 3)(4x + 1)
29.
7p + 16q
56
40.
30.
33t + 21
77
41. x = 8
31.
10x + 39
(x − 1)(x + 6)
33.
5a + 3
28
34.
9pq
56
35.
42. x = 5
43. x = 2
y
32.
72
44. x = 7
45. x = 12
46. x = 3
47. x = 9
pq + p
7q
48. x = 5
Answers for Section 3
√
1. 4 2
√
2. 5 7
√
3. 4 11
√
4. 7 3
A.4
33
35
√
5. 8 2
√
6. 8 5
√
7. 3 3
√
8. 3 2
9. 6
√
10. 3 10
√
11. 14 5
√
12. 15 14
13. 2
√
14. 6 + 4 2
√
15. 12 − 2 35
√
16. 3 2
Answers for Section 4
1. 511
7. c2
13. 211
2. 69
8. d−4
14. 75
3. 35
9. 320
15. 53
4. 7−2
10. x14
16. 1
5. a5
11. y 15
17. 53 2
6. b−3
12. z 14
18. 33 3
1
2
31
1
33. a = − 41
19. 23 2
1
20. 7−1 2
34. b =
4
21. 7 3
2
22. a1 5
2
23. b4 3
1
24. c1 6
5
25. x 6
1
26. x5 3
47.
5
3
48. 4 15
1
6
35. c = −5
49. a = 25
1
36. d = − 10
50. b = 81
37. x = 2
51. c = 243
38. y = 5
52. d =
39. z = 1
53. log3 (x2 + 4x − 5)
40. x =
1
36
54. log2 (x4 + 5x3 )
4
9
41. 5
)
55. log5 ( 2x+7
7x+2
28. x−2 2
42. 5
56. log5 ( x
29. x = 9
43. 2 12
57. log3 (162x − 405)
30. x = 2
44. −2
58. log4 (6x + 10)
31. x = 1 21
45.
5
27. x− 6
1
32. x =
A.5
1
3
1
2
x−9
)
59. log6 ( 12 x + 13 )
46. − 12
60. log9 (x3 + 3x2 − 9x − 27)
Answers for Section 5
√
22. − 3
1
2
1. − sin(φ)
12.
2. sin(φ)
13. 0
3. − sin(φ)
14.
1
2
15.
1
2
25.
1
π
2
√
16. − 12 3
26.
3
π
4
17. −1
27.
1
π
4
18. − 12
√
19. 3
28.
1
π
3
29.
1
4
4. cos(φ)
5. cos(φ)
6. cos(φ)
7. − cos(φ)
8. − cos(φ)
9. sin(φ)
10. − sin(a)
√
11. 21 3
A.6
2 +6x+9
23.
√
3
1
π
6
24. − 12 π
√
√
3)
√
2(1 + 3)
p
√
31. 12 2 − 2
√
20. − 13 3
30.
21. 0
Answers for Section 6
32
1
4
√
2(1 +
√
√
5− 57
5+ 57
or
x
=
4
4
√
√
−1+ 21
−1− 21
or
x
=
10
10
1. a = 3
30. x =
2. b = 11
31. x =
3. c = 2
32. No solution as b2 − 4ac < 0
4. d = −1
33. x = 0 or x = 3 or x = 4
5. e = 5
34. x = 0 or x = − 75
6. f = −3
35. x = 0 or x = −2 or x = 2
7. g = 2 31
36. x = −2 or x = 2 or x = −1 or x = 1
8. g = − 71
37. x = 0 or x = −2 or x = 2 or x = −1 or
x=1
9. h = 5
38. x = −3 or x = 3
10. k = −2
11. m =
9
− 19
12. n =
2
11
39. x = 0 or x = −3 or x =
40. x =
42. x = −2 or x = 1 51
14. x = −2 or x = −5
43. x = −7 or x = −3 or x = 3 or x = 5
15. x = 1 or x = −5
44. x = − 75 or x = 0 or x = 1 or x = 2
√
45. x = 5 or x = 0 or x = 3 5
16. x = 6 or x = 3
17. x = −1 or x = 3
46. x = −1 or x = 0 or x = 5
18. x = −7 or x = 11
47. x = −4 or x =
19. x = −2 or x = −5
2
5
48. x = −3 or x = 3
20. x = 4
49. x = 7
21. x = 1 or x = 2
50. x = 3
22. x = −3 or x = 1
51. x = 0 or x = 4
23. x = −2 or x = 2
52. x = 4
24. x = −9 or x = 1
53. x = 1
2
3
54. x = 1 or x = 3
26. x = −2 21 or x = −1 21
27. x = −1 or x =
3
4
28. x = −4 or x = − 51
29. x =
or x = − 21
41. x = −4 or x = −2
13. x = 7 or x = 3
25. x = −3 or
1
2
1
4
√
−8−2 10
6
or x =
√
−8+2 10
6
55. x = 5
56. x = 1 or x = 4
√
57. a = 3 17
√
√
58. b = − 3 7 or b = 3 7
33
59. c =
√
3
16
1
60. d = 11− 3 =
68. x = 7
1
√
3
11
61. x = 3 + log5 (123)
62. x = 3 + 21 log2 (12)
63. x = − 13 + 13 log3 (12)
64. x = 3 + log5 (24)
69. x = 16 π + k2π or x = 56 π + k2π
70. x = 16 π + kπ or x = 13 π + kπ
71. x = 34 π + k2π or x = 54 π + k2π
72. x = 16 π + k 23 π or x =
73. x = 14 π + kπ
65. x = 5
1
π
12
+ k 12 π
66. x = 3 21
74. x =
67. x = 4
75. x = 14 π + k 12 π
A.7
Answers for Section 7
1
1. p = (q 2 − 1)
2
2. A =
r
5. v =
1
2(E − mgh)
m
6. P = 2I
B2C 3
3. R =
Q
I 2t
7. x =
2y
1−y
4. m =
2E
2gh + v 2
8. t =
y2 − 1
2y
A.8
Answers for Section 8
1. f 0 (x) = 18x2 + 6x + 4
2. f 0 (x) = 20x4 − 15x2 − 2x + 7
3. f 0 (x) = 15x4 + 16x3
4. f 0 (x) = 16x3 − 24x
12
3
−
x2 x4
√
3
4
6. f 0 (x) = 10x x + √ − √
x x x
5. f 0 (x) = −
7. f 0 (x) = 4x3 −
4
x5
8. f 0 (x) = 4x3 − 3x2 − 12x
9. f 0 (x) = 4ex
10. f 0 (x) = 3x ln(3) + 5x ln(5)
34
11
π
18
+ k 23 π
1
1
11. f 0 (x) = ( )x ln( ) = −2−x ln(2)
2
2
12. f 0 (x) = 3ex + e3x ln(3) + 3e
13. f 0 (x) = 35(x2 − 2x)6 (2x − 2)
14. f 0 (x) =
−3x2 − 5
(x3 + 5x − 1)2
6x
15. f 0 (x) = √
x2 + 1
16. f 0 (x) = ex
17. f 0 (x) =
18. f 0 (x) =
3 +x
(3x2 + 1)
−6x − 9
(x2 + 3x + 1)4
(x2
19. f 0 (x) = −
−x
√
+ 1) x2 + 1
e1/x
x2
1 + ex
20. f 0 (x) = √
2 x + ex
√
e x
21. f (x) = √
2 x
0
22. f 0 (x) =
1√ x
e
2
23. f 0 (x) =
−2e2x − 2x
(e2x + x2 )2
x
3x2 + 2√e e+1
x +x
24. f (x) = − 3 √ x
(x + e + x)2
0
25. f 0 (x) = (9x2 + 8x)(2x4 + 3x + 5) + (3x3 + 4x2 + 1)(8x3 + 3)
26. f 0 (x) = (x + 5)7 + 7x(x + 5)6
27. f 0 (x) = 3x2 (x3 − 3x2 + 6)5 + 5x3 (x3 − 3x2 + 6)4 (3x2 − 6x)
√
2x5
28. f 0 (x) = 5x4 4x − 1 + √
4x − 1
29. f 0 (x) = ex + xex
30. f 0 (x) = ex (x3 + x2 + x + 1) + ex (3x2 + 2x + 1)
√
x2 (1 + ex )
31. f 0 (x) = 2x x + ex + √
2 x + ex
35
√
x+1
32. f 0 (x) = ex x2 + 2x + ex √
x2 + 2x
33. f 0 (x) =
2x(2x + 1) − 2x2
(2x + 1)2
34. f 0 (x) =
2x(x2 − 1) − 2x(x2 + 1)
(x2 − 1)2
35. f 0 (x) =
(6x2 − 6x)(x + 3) − (2x3 − 3x2 )
(x + 3)2
36. f 0 (x) =
3x2 (x2 − 2) − (x3 − 3)(2x)
(x2 − 2)2
37. f 0 (x) =
ex (ex + 1) − ex · ex
(ex + 1)2
(3x2 (2x + 1)2 + x3 · 4(2x + 1))(x3 − 1) − 3x5 (2x + 1)2
38. f (x) =
(x3 − 1)2
0
6(2x + 5)2 ex − (2x + 5)3 ex
39. f (x) =
e2x
√
√
(1 + 2√1 x )(1 + x) − (x + x)( 2√1 x )
0
√
40. f (x) =
(1 + x)2
0
A.9
Answers for Section 9
1. y = 17x − 66
2. y = 9x − 12
3. y = 35 x +
6
5
4. y = 15x − 26
5. y = −3
6. y = 6x − 8
7. y = 3ex − 2e
8. y = −125x
9. Minimum f (4) = −5
10. Maximum f (2) = 10 23 , minimum f (6) = 0
11. Minimum f (−5) = 0, maximum f (−4) = 16, minimum f (0) = 0
12. Minimum f (−1) = 43 , maximum f (3) =
1
12
13. (−1, −13) and (1, −11)
36
14. (−1, 14
) and (−3, −2e3 )
e
√
15. (2, 8 2)
16. (0, 0), (−3, − 14 ), and (3, 41 )
37