# MTH 111–112 Supplement

## AppendixAAnswers and Solutions to Exercises

### 1MTH 111 Supplement1.1Graph Transformations

#### Exercises

##### 1.1.1.
$$g(x)=f(-x)\text{.}$$ So, we can reflect the graph of $$y=f(x)$$ across the $$y$$-axis to obtain $$y=g(x)\text{.}$$
##### 1.1.2.
$$h(x)=-f(x)\text{.}$$ So, we can reflect the graph of $$y=f(x)$$ across the $$x$$-axis to obtain $$y=h(x)\text{.}$$
##### 1.1.3.
$$k(x)=f(x)+6\text{.}$$ So, we can shift the graph of $$y=f(x)$$ up $$6$$ units to obtain $$y=k(x)\text{.}$$
##### 1.1.4.
$$l(x)=3f(x)\text{.}$$ So, we can stretch the graph of $$y=f(x)$$ vertically by a factor of $$3$$ to obtain $$y=l(x)\text{.}$$
##### 1.1.5.
 $$x$$ $$-4$$ $$-3$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$f(x)$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$\frac{1}{2}x$$ $$-1$$ $$-\frac{1}{2}$$ $$0$$ $$\frac{1}{2}$$ $$1$$ $$\frac{3}{2}$$ $$2$$ $$\frac{5}{2}$$ $$3$$ $$-2f(x)$$ $$4$$ $$2$$ $$0$$ $$-2$$ $$-4$$ $$-6$$ $$-8$$ $$-10$$ $$-12$$ $$f(x)+5$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$f(x+2)$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$f\mathopen{}\left( \frac{1}{2}x \right)\mathclose{}$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$f(2x)$$ $$-2$$ $$0$$ $$2$$ $$4$$ $$6$$ $$f(x-3)$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$
##### 1.1.6.
Solution.
\begin{align*} g(x) \amp = \frac{\sqrt{x-7}}{4} \\ \amp = \frac{1}{4}\sqrt{x-7} \\ \amp = \frac{1}{4}f(x-7) \end{align*}
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 $$\frac{1}{4}\text{.}$$ (There are other correct answers.)
##### 1.1.7.
Solution.
\begin{align*} g(x) \amp = \frac{2}{x}+3 \\ \amp = 2 \cdot \frac{1}{x}+3 \\ \amp = 2f(x)+3 \end{align*}
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{align*} g(x) \amp =-4\left( \frac{1}{2}x-5 \right)^2+3 \\ \amp = -4f\mathopen{}\left( \frac{1}{2}x-5 \right)\mathclose{}+3 \\ \amp = -4f\mathopen{}\left( \frac{1}{2} (x-10) \right)\mathclose{}+3 \end{align*}
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{align*} g(x) \amp =\frac{1}{2}\sqrt[3]{10x+30}-6 \\ \amp =\frac{1}{2}f(10x+30)-6 \\ \amp =\frac{1}{2}f\mathopen{}\left( 10( x+3 ) \right)\mathclose{} -6 \end{align*}
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.)

### 1.2Inverse 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 $$m^{-1}\text{.}$$
 $$x$$ $$0$$ $$5$$ $$10$$ $$15$$ $$20$$ $$m^{-1}(x)$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
##### 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.

### 1.3Exponential Functions

#### Exercises

##### 1.3.1.
$$f(x)=50 \cdot 2^x$$
##### 1.3.2.
$$f(x)=4 \cdot \left( \frac{1}{2} \right)^x$$
##### 1.3.3.
$$f(x)=2 \cdot 3^x$$
##### 1.3.4.
$$f(x)=10 \cdot \left( \frac{4}{5} \right)^x$$
##### 1.3.5.
$$f(x)=5 \cdot \left( \frac{1}{5} \right)^x$$
##### 1.3.6.
$$f(x)=\frac{1}{2} \cdot \left( \frac{2}{3} \right)^x$$

### 1.4Logarithmic Functions

#### Exercises

##### 1.4.1.
$$a=\sqrt{5}$$
##### 1.4.2.
$$a=20$$
##### 1.4.3.
$$a=\sqrt{3}$$

### 2MTH 112 Supplement2.1Angles2.1.4Exercises

#### 2.1.4.1.

$$423^{\circ}$$ and $$-297^{\circ}$$ are coterminal with $$63^{\circ}\text{.}$$

#### 2.1.4.2.

$$\frac{19\pi}{9}$$ and $$-\frac{17\pi}{9}$$ are coterminal with $$\frac{\pi}{9}\text{.}$$

#### 2.1.4.3.

$$\frac{29\pi}{8}$$ and $$-\frac{3\pi}{8}$$ are coterminal with $$\frac{13\pi}{8}\text{.}$$

#### 2.1.4.4.

$$60^{\circ}$$

#### 2.1.4.5.

$$\frac{\pi}{4}$$

#### 2.1.4.6.

$$40^{\circ}$$

#### 2.1.4.7.

$$\frac{3\pi}{8}$$

#### 2.1.4.8.

$$\pi-2 \approx 1.14$$

#### 2.1.4.9.

$$\frac{\pi}{11}$$

#### 2.1.4.10.

$$20^{\circ}$$

#### 2.1.4.11.

$$\frac{\pi}{5}$$

#### 2.1.4.12.

$$80^{\circ}$$

#### 2.1.4.13.

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

#### 2.1.4.14.

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

#### 2.1.4.15.

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

#### 2.1.4.16.

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

#### 2.1.4.17.

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

#### 2.1.4.18.

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

#### 2.1.4.19.

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

#### 2.1.4.20.

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

### 2.2Generalized Definitions of Trigonometric Functions

#### Exercises

##### 2.2.1.
\begin{align*} \cos(\theta) \amp= \frac{3}{5} \\ \sin(\theta) \amp= \frac{4}{5} \\ \tan(\theta) \amp= \frac{4}{3} \\ \sec(\theta) \amp= \frac{5}{3} \\ \csc(\theta) \amp= \frac{5}{4} \\ \cot(\theta) \amp= \frac{3}{4} \end{align*}
##### 2.2.2.
\begin{align*} \sin(\theta) \amp= -\frac{\sqrt{10}}{10} \\ \cos(\theta) \amp= -\frac{3\sqrt{10}}{10} \\ \tan(\theta) \amp= 3 \\ \sec(\theta) \amp= -\sqrt{10} \\ \csc(\theta) \amp= -\frac{\sqrt{10}}{3} \\ \cot(\theta) \amp= \frac{1}{3} \end{align*}
##### 2.2.3.
$$\mathopen{}\left( -\frac{3\sqrt{3}}{2}, -\frac{3}{2} \right)\mathclose{}$$
##### 2.2.4.
$$\mathopen{}\left( 5, 5\sqrt{3} \right)\mathclose{}$$

### 2.3Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

#### Exercises

##### 2.3.1.
Amplitude
$$3$$ units
Period
$$\frac{2\pi}{3}$$ units
Midline
$$\displaystyle y=0$$
Phase shift
$$\displaystyle \frac{\pi}{2}$$
Horizontal Shift
$$\frac{\pi}{6}$$ units to the right
##### 2.3.2.
Amplitude
$$1$$ unit
Period
$$\frac{\pi}{2}$$ units
Midline
$$\displaystyle y=3$$
Phase shift
$$\displaystyle -\pi$$
Horizontal Shift
$$\frac{\pi}{4}$$ units to the left
##### 2.3.3.
Amplitude
$$2$$ units
Period
$$1$$ unit
Midline
$$\displaystyle y=4$$
Phase shift
$$\displaystyle \pi$$
Horizontal Shift
$$\frac{1}{2}$$ of a unit to the right
##### 2.3.4.
Amplitude
$$4$$ units
Period
$$2$$ units
Midline
$$\displaystyle y=-2$$
Phase shift
$$\displaystyle -\frac{\pi}{4}$$
Horizontal Shift
$$\frac{1}{4}$$ of a unit to the left
##### 2.3.5.
\begin{aligned} p(x)=4 \sin \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{4} \right)\mathclose{} \right)\mathclose{}-2 \\ p(x)=4 \cos \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{2} \right)\mathclose{} \right)\mathclose{}-2 \end{aligned}
##### 2.3.6.
\begin{aligned}q(x)=3 \sin \mathopen{}\left( \pi \mathopen{}\left( x+\frac{1}{4} \right)\mathclose{} \right)\mathclose{}-1\\ q(x)=3 \cos \mathopen{}\left( \pi \mathopen{}\left( x-\frac{1}{4} \right)\mathclose{} \right)\mathclose{}-1\end{aligned}

### 2.4Complex Numbers and Polar Coordinates2.4.3Exercises

#### 2.4.3.1.

$$z=12e^{\frac{\pi}{3} \cdot i}$$

#### 2.4.3.2.

$$z=4e^{\frac{5\pi}{6} \cdot i}$$

#### 2.4.3.3.

$$z=10e^{-\frac{\pi}{4} \cdot i}$$

#### 2.4.3.4.

$$z=4\sqrt{3}+4i$$

#### 2.4.3.5.

$$z=-4$$

#### 2.4.3.6.

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

#### 2.4.3.7.

$$\sqrt{18-18\sqrt{3}\cdot i}=3\sqrt{3}-3i$$ (and the non-principal root is $$-3\sqrt{3}+3i$$)

#### 2.4.3.8.

$$\sqrt[3]{-16+16i}=2+2i$$ (and the non-principal roots are $$\left(-1-\sqrt{3}\right)+\left(-1+\sqrt{3}\right)i$$ and $$\left(-1+\sqrt{3}\right)+\left(-1-\sqrt{3}\right)i$$)

#### 2.4.3.9.

$$\sqrt{-i} = \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$$ (and the non-principal root is $$-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$)

#### 2.4.3.10.

$$\sqrt[5]{-16\sqrt{3}-16i}=\sqrt{3}-i$$ (and the non-principal roots are $$2\cos\mathopen{}\left(\frac{7\pi}{30}\right)+2i\sin\mathopen{}\left(\frac{7\pi}{30}\right)\text{,}$$ $$2\cos\mathopen{}\left(\frac{19\pi}{30}\right)+2i\sin\mathopen{}\left(\frac{19\pi}{30}\right)\text{,}$$ $$2\cos\mathopen{}\left(\frac{29\pi}{30}\right)-2i\sin\mathopen{}\left(\frac{29\pi}{30}\right)\text{,}$$ and $$2\cos\mathopen{}\left(\frac{17\pi}{30}\right)-2i\sin\mathopen{}\left(\frac{17\pi}{30}\right)$$)

#### 2.4.3.11.

$$\frac{3\sqrt{3}}{2}+\frac{3}{2}i\text{,}$$ $$-3i\text{,}$$ and $$-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
$$\frac{1}{2}+\frac{\sqrt{3}}{2}i$$ and $$-\frac{1}{2}-\frac{\sqrt{3}}{2}i$$
$$\left\{ -1, \frac{1}{2}+\frac{\sqrt{3}}{2}i, \frac{1}{2}-\frac{\sqrt{3}}{2}i \right\}$$