Answers and Solutions to Exercises

Appendix A Answers and Solutions to Exercises

1 MTH 111 Supplement
1.1 Graph Transformations

Exercises

1.1.1.

Answer.

g (x) = f (- x) .

So, we can reflect the graph of

y = f (x)

across the

y

-axis to obtain

y = g (x) .

1.1.2.

Answer.

h (x) = - f (x) .

So, we can reflect the graph of

y = f (x)

across the

x

-axis to obtain

y = h (x) .

1.1.3.

Answer.

k (x) = f (x) + 6 .

So, we can shift the graph of

y = f (x)

6

units to obtain

y = k (x) .

1.1.4.

Answer.

l (x) = 3 f (x) .

So, we can stretch the graph of

y = f (x)

vertically by a factor of

3

to obtain

y = l (x) .

1.1.5.

Answer.

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$	$4$
$f (x)$	$- 2$	$- 1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$\frac{1}{2} x$	$- 1$	$- \frac{1}{2}$	$0$	$\frac{1}{2}$	$1$	$\frac{3}{2}$	$2$	$\frac{5}{2}$	$3$
$- 2 f (x)$	$4$	$2$	$0$	$- 2$	$- 4$	$- 6$	$- 8$	$- 10$	$- 12$
$f (x) + 5$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$
$f (x + 2)$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$f (\frac{1}{2} x)$	$0$		$1$		$2$		$3$		$4$
$f (2 x)$			$- 2$	$0$	$2$	$4$	$6$
$f (x - 3)$				$- 2$	$- 1$	$0$	$1$	$2$	$3$

1.1.6.

Solution.

\begin{aligned} g (x) & = \frac{\sqrt{x - 7}}{4} \\ = \frac{1}{4} \sqrt{x - 7} \\ = \frac{1}{4} f (x - 7) \end{aligned}

So we can transform

y = f (x)

into

y = g (x)

by first shifting right

7

units and then compressing vertically by a factor of

\frac{1}{4} .

(There are other correct answers.)

1.1.7.

Solution.

\begin{aligned} g (x) & = \frac{2}{x} + 3 \\ = 2 \cdot \frac{1}{x} + 3 \\ = 2 f (x) + 3 \end{aligned}

So we can transform

y = f (x)

into

y = g (x)

by first stretching vertically by a factor of

2

and then shifting up

3

units. (There are other correct answers.)

1.1.8.

Solution.

\begin{aligned} g (x) & = - 4 {(\frac{1}{2} x - 5)}^{2} + 3 \\ = - 4 f (\frac{1}{2} x - 5) + 3 \\ = - 4 f (\frac{1}{2} (x - 10)) + 3 \end{aligned}

So we can transform

y = f (x)

into

y = g (x)

by first stretching horizontally by a factor of

2

and then shifting right

10

units. Then, stretching vertically by a factor of

4

and reflecting across the

x

-axis, and finally shifting up

3

units. (There are other correct answers.)

1.1.9.

Solution.

\begin{aligned} g (x) & = \frac{1}{2} \sqrt[3]{10 x + 30} - 6 \\ = \frac{1}{2} f (10 x + 30) - 6 \\ = \frac{1}{2} f (10 (x + 3)) - 6 \end{aligned}

So we can transform

y = f (x)

into

y = g (x)

by first compressing horizontally by a factor of

\frac{1}{10}

and then shifting left

3

units. Then, compressing vertically by a factor of

\frac{1}{2}

and finally shifting down

6

units. (There are other correct answers.)

1.1.10.

Answer.

$The graph has the points (-2,4), (-1, 0), (0, 4), (1, 4), (2, -4) all connected.$

1.1.11.

Answer.

$The graph has the points (-2, -9), (-1, 7), (0, 7), (1, -1), and (2, 7) connected.$

1.1.12.

Answer.

$This graph has the points (-4, -8), (-3, 0), (-2, -8), (-1, -8), and (0, 8) all connected.$

1.1.13.

Answer.

$This graph has the points (-8, 6), (-4, 2), (0, 6), (4, 6), and (8, -2) are all connected.$

1.2 Inverse Functions

Exercises

1.2.1.

Solution.

m

is an invertible function since it is one-to-one, i.e., each output corresponds to exactly one input. Here is a table-of-values for

m^{- 1} .

$x$	$0$	$5$	$10$	$15$	$20$
$m^{- 1} (x)$	$1$	$2$	$3$	$4$	$5$

1.2.2.

Solution.

p

isn’t an invertible function since it isn’t one-to-one. Notice how the output

0

corresponds to two distinct output values.

1.3 Exponential Functions

Exercises

1.3.1.

Answer.

f (x) = 50 \cdot 2^{x}

1.3.2.

Answer.

f (x) = 4 \cdot {(\frac{1}{2})}^{x}

1.3.3.

Answer.

f (x) = 2 \cdot 3^{x}

1.3.4.

Answer.

f (x) = 10 \cdot {(\frac{4}{5})}^{x}

1.3.5.

Answer.

f (x) = 5 \cdot {(\frac{1}{5})}^{x}

1.3.6.

Answer.

f (x) = \frac{1}{2} \cdot {(\frac{2}{3})}^{x}

1.4 Logarithmic Functions

Exercises

1.4.1.

Answer.

a = \sqrt{5}

1.4.2.

Answer.

a = 20

1.4.3.

Answer.

a = \sqrt{3}

2 MTH 112 Supplement
2.1 Angles
2.1.4 Exercises

2.1.4.1.

Answer.

423^{\circ}

and

- 297^{\circ}

are coterminal with

63^{\circ} .

2.1.4.2.

Answer.

\frac{19 π}{9}

and

- \frac{17 π}{9}

are coterminal with

\frac{π}{9} .

2.1.4.3.

Answer.

\frac{29 π}{8}

and

- \frac{3 π}{8}

are coterminal with

\frac{13 π}{8} .

2.1.4.4.

Answer.

60^{\circ}

2.1.4.5.

Answer.

\frac{π}{4}

2.1.4.6.

Answer.

40^{\circ}

2.1.4.7.

Answer.

\frac{3 π}{8}

2.1.4.8.

Answer.

π - 2 \approx 1.14

2.1.4.9.

Answer.

\frac{π}{11}

2.1.4.10.

Answer.

20^{\circ}

2.1.4.11.

Answer.

\frac{π}{5}

2.1.4.12.

Answer.

80^{\circ}

2.1.4.13.

Answer.

243^{\circ} 10^{'} \approx {243.167}^{\circ}

2.1.4.14.

Answer.

3^{\circ} 25^{'} \approx {3.417}^{\circ}

2.1.4.15.

Answer.

- 23^{\circ} 3^{'} = - {23.05}^{\circ}

2.1.4.16.

Answer.

75^{\circ} 32^{'} 17^{″} \approx {75.538}^{\circ}

2.1.4.17.

Answer.

{12.4}^{\circ} = 12^{\circ} 24^{'}

2.1.4.18.

Answer.

{1.53}^{\circ} = 1^{\circ} 31^{'} 48^{″}

2.1.4.19.

Answer.

- {144.9}^{\circ} = - 144^{\circ} 54^{'}

2.1.4.20.

Answer.

{0.416}^{\circ} = 0^{\circ} 24^{'} {57.6}^{″}

2.2 Generalized Definitions of Trigonometric Functions

Exercises

2.2.1.

Answer.

\begin{aligned} \cos (θ) & = \frac{3}{5} \\ \sin (θ) & = \frac{4}{5} \\ \tan (θ) & = \frac{4}{3} \\ \sec (θ) & = \frac{5}{3} \\ \csc (θ) & = \frac{5}{4} \\ \cot (θ) & = \frac{3}{4} \end{aligned}

2.2.2.

Answer.

\begin{aligned} \sin (θ) & = - \frac{\sqrt{10}}{10} \\ \cos (θ) & = - \frac{3 \sqrt{10}}{10} \\ \tan (θ) & = 3 \\ \sec (θ) & = - \sqrt{10} \\ \csc (θ) & = - \frac{\sqrt{10}}{3} \\ \cot (θ) & = \frac{1}{3} \end{aligned}

2.2.3.

Answer.

(- \frac{3 \sqrt{3}}{2}, - \frac{3}{2})

2.2.4.

Answer.

(5, 5 \sqrt{3})

2.3 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Exercises

2.3.1.

Answer.

Amplitude: $3$ units
Period: $\frac{2 π}{3}$ units
Midline: $y = 0$
Phase shift: $\frac{π}{2}$
Horizontal Shift: $\frac{π}{6}$ units to the right

$The graph of f(x)=3sin(3x-pi/2).$

2.3.2.

Answer.

Amplitude: $1$ unit
Period: $\frac{π}{2}$ units
Midline: $y = 3$
Phase shift: $- π$
Horizontal Shift: $\frac{π}{4}$ units to the left

$The graph of g(t)=cos(4t+pi)+3.$

2.3.3.

Answer.

Amplitude: $2$ units
Period: $1$ unit
Midline: $y = 4$
Phase shift: $π$
Horizontal Shift: $\frac{1}{2}$ of a unit to the right

$The graph of m(theta)=2cos(2 pi theta - pi) + 4.$

2.3.4.

Answer.

Amplitude: $4$ units
Period: $2$ units
Midline: $y = - 2$
Phase shift: $- \frac{π}{4}$
Horizontal Shift: $\frac{1}{4}$ of a unit to the left

$The graph of n(x)=-4sin(pi x + pi/4)-2.$

2.3.5.

Answer.

\begin{array}{r} p (x) = 4 \sin (2 (x - \frac{π}{4})) - 2 \\ p (x) = 4 \cos (2 (x - \frac{π}{2})) - 2 \end{array}

2.3.6.

Answer.

\begin{array}{r} q (x) = 3 \sin (π (x + \frac{1}{4})) - 1 \\ q (x) = 3 \cos (π (x - \frac{1}{4})) - 1 \end{array}

2.4 Complex Numbers and Polar Coordinates
2.4.3 Exercises

2.4.3.1.

Answer.

z = 12 e^{\frac{π}{3} \cdot i}

2.4.3.2.

Answer.

z = 4 e^{\frac{5 π}{6} \cdot i}

2.4.3.3.

Answer.

z = 10 e^{- \frac{π}{4} \cdot i}

2.4.3.4.

Answer.

z = 4 \sqrt{3} + 4 i

2.4.3.5.

Answer.

z = - 4

2.4.3.6.

Answer.

z = - \frac{5}{2} - \frac{5 \sqrt{3}}{2} \cdot i

2.4.3.7.

Answer.

\sqrt{18 - 18 \sqrt{3} \cdot i} = 3 \sqrt{3} - 3 i

(and the non-principal root is

- 3 \sqrt{3} + 3 i

)

2.4.3.8.

Answer.

\sqrt[3]{- 16 + 16 i} = 2 + 2 i

(and the non-principal roots are

(- 1 - \sqrt{3}) + (- 1 + \sqrt{3}) i

and

(- 1 + \sqrt{3}) + (- 1 - \sqrt{3}) i

)

2.4.3.9.

Answer.

\sqrt{- i} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i

(and the non-principal root is

- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i

)

2.4.3.10.

Answer.

\sqrt[5]{- 16 \sqrt{3} - 16 i} = \sqrt{3} - i

(and the non-principal roots are

2 \cos (\frac{7 π}{30}) + 2 i \sin (\frac{7 π}{30}),

2 \cos (\frac{19 π}{30}) + 2 i \sin (\frac{19 π}{30}),

2 \cos (\frac{29 π}{30}) - 2 i \sin (\frac{29 π}{30}),

and

2 \cos (\frac{17 π}{30}) - 2 i \sin (\frac{17 π}{30})

)

2.4.3.11.

Answer.

\frac{3 \sqrt{3}}{2} + \frac{3}{2} i,

- 3 i,

and

- \frac{3 \sqrt{3}}{2} + \frac{3}{2} i

2.4.3.12.

Answer.

\frac{1}{2} + \frac{\sqrt{3}}{2} i

and

- \frac{1}{2} - \frac{\sqrt{3}}{2} i

2.4.3.13.

Answer.

{- 1, \frac{1}{2} + \frac{\sqrt{3}}{2} i, \frac{1}{2} - \frac{\sqrt{3}}{2} i}

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Appendix A Answers and Solutions to Exercises

1 MTH 111 Supplement1.1 Graph Transformations

Exercises

1.1.1.

1.1.2.

1.1.3.

1.1.4.

1.1.5.

1.1.6.

1.1.7.

1.1.8.

1.1.9.

1.1.10.

1.1.11.

1.1.12.

1.1.13.

1.2 Inverse Functions

Exercises

1.2.1.

1.2.2.

1.3 Exponential Functions

Exercises

1.3.1.

1.3.2.

1.3.3.

1.3.4.

1.3.5.

1.3.6.

1.4 Logarithmic Functions

Exercises

1.4.1.

1.4.2.

1.4.3.

2 MTH 112 Supplement2.1 Angles2.1.4 Exercises

2.1.4.1.

2.1.4.2.

2.1.4.3.

2.1.4.4.

2.1.4.5.

2.1.4.6.

2.1.4.7.

2.1.4.8.

2.1.4.9.

2.1.4.10.

2.1.4.11.

2.1.4.12.

2.1.4.13.

2.1.4.14.

2.1.4.15.

2.1.4.16.

2.1.4.17.

2.1.4.18.

2.1.4.19.

2.1.4.20.

2.2 Generalized Definitions of Trigonometric Functions

Exercises

2.2.1.

2.2.2.

2.2.3.

2.2.4.

2.3 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Exercises

2.3.1.

2.3.2.

2.3.3.

2.3.4.

2.3.5.

2.3.6.

2.4 Complex Numbers and Polar Coordinates2.4.3 Exercises

2.4.3.1.

2.4.3.2.

2.4.3.3.

2.4.3.4.

2.4.3.5.

2.4.3.6.

2.4.3.7.

2.4.3.8.

2.4.3.9.

2.4.3.10.

2.4.3.11.

1 MTH 111 Supplement
1.1 Graph Transformations

2 MTH 112 Supplement
2.1 Angles
2.1.4 Exercises

2.4 Complex Numbers and Polar Coordinates
2.4.3 Exercises