1.1.1.
Answer.
\(g(x)=f(-x)\text{.}\) So, we can reflect the graph of \(y=f(x)\) across the \(y\)-axis to obtain \(y=g(x)\text{.}\)
Appendix A Answers and Solutions to Exercises
1 MTH 111 Supplement
1.1 Graph Transformations
Exercises
1.1.2.
Answer.
\(h(x)=-f(x)\text{.}\) So, we can reflect the graph of \(y=f(x)\) across the \(x\)-axis to obtain \(y=h(x)\text{.}\)
1.1.3.
Answer.
\(k(x)=f(x)+6\text{.}\) So, we can shift the graph of \(y=f(x)\) up \(6\) units to obtain \(y=k(x)\text{.}\)
1.1.4.
Answer.
\(l(x)=3f(x)\text{.}\) 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.
Answer.
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
| \(f(x)\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
| \(\frac{1}{2}x\) | \(-1\) | \(-\frac{1}{2}\) | \(0\) | \(\frac{1}{2}\) | \(1\) | \(\frac{3}{2}\) | \(2\) | \(\frac{5}{2}\) | \(3\) |
| \(-2f(x)\) | \(4\) | \(2\) | \(0\) | \(-2\) | \(-4\) | \(-6\) | \(-8\) | \(-10\) | \(-12\) |
| \(f(x)+5\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) |
| \(f(x+2)\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | ||
| \(f\mathopen{}\left( \frac{1}{2}x \right)\mathclose{}\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | ||||
| \(f(2x)\) | \(-2\) | \(0\) | \(2\) | \(4\) | \(6\) | ||||
| \(f(x-3)\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
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.)
1.1.10.
Answer.
1.1.11.
Answer.
1.1.12.
Answer.
1.1.13.
Answer.
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 \(m^{-1}\text{.}\)
| \(x\) | \(0\) | \(5\) | \(10\) | \(15\) | \(20\) |
| \(m^{-1}(x)\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
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.
1.3 Exponential Functions
Exercises
1.3.1.
Answer.
\(f(x)=50 \cdot 2^x\)
1.3.2.
Answer.
\(f(x)=4 \cdot \left( \frac{1}{2} \right)^x\)
1.3.3.
Answer.
\(f(x)=2 \cdot 3^x\)
1.3.4.
Answer.
\(f(x)=10 \cdot \left( \frac{4}{5} \right)^x\)
1.3.5.
Answer.
\(f(x)=5 \cdot \left( \frac{1}{5} \right)^x\)
1.3.6.
Answer.
\(f(x)=\frac{1}{2} \cdot \left( \frac{2}{3} \right)^x\)
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 \(63^{\circ}\text{.}\)
2.1.4.2.
Answer.
\(\frac{19\pi}{9}\) and \(-\frac{17\pi}{9}\) are coterminal with \(\frac{\pi}{9}\text{.}\)
2.1.4.3.
Answer.
\(\frac{29\pi}{8}\) and \(-\frac{3\pi}{8}\) are coterminal with \(\frac{13\pi}{8}\text{.}\)
2.1.4.4.
Answer.
\(60^{\circ}\)
2.1.4.5.
Answer.
\(\frac{\pi}{4}\)
2.1.4.6.
Answer.
\(40^{\circ}\)
2.1.4.7.
Answer.
\(\frac{3\pi}{8}\)
2.1.4.8.
Answer.
\(\pi-2 \approx 1.14\)
2.1.4.9.
Answer.
\(\frac{\pi}{11}\)
2.1.4.10.
Answer.
\(20^{\circ}\)
2.1.4.11.
Answer.
\(\frac{\pi}{5}\)
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{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.
Answer.
\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.
Answer.
\(\mathopen{}\left( -\frac{3\sqrt{3}}{2}, -\frac{3}{2} \right)\mathclose{}\)
2.2.4.
Answer.
\(\mathopen{}\left( 5, 5\sqrt{3} \right)\mathclose{}\)
2.3 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift
Exercises
2.3.1.
Answer.
- 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.
Answer.
- 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.
Answer.
- 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.
Answer.
- 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.
Answer.
\(\begin{aligned} p(x)=4 \sin \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{4} \right)\mathclose{} \right)\mathclose{}-2 \\ p(x)=4 \cos \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{2} \right)\mathclose{} \right)\mathclose{}-2 \end{aligned} \)
2.3.6.
Answer.
\(\begin{aligned}q(x)=3 \sin \mathopen{}\left( \pi \mathopen{}\left( x+\frac{1}{4} \right)\mathclose{} \right)\mathclose{}-1\\ q(x)=3 \cos \mathopen{}\left( \pi \mathopen{}\left( x-\frac{1}{4} \right)\mathclose{} \right)\mathclose{}-1\end{aligned}\)
2.4 Complex Numbers and Polar Coordinates
2.4.3 Exercises
2.4.3.1.
Answer.
\(z=12e^{\frac{\pi}{3} \cdot i}\)
2.4.3.2.
Answer.
\(z=4e^{\frac{5\pi}{6} \cdot i}\)
2.4.3.3.
Answer.
\(z=10e^{-\frac{\pi}{4} \cdot i}\)
2.4.3.4.
Answer.
\(z=4\sqrt{3}+4i\)
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.
\(\sqrt{18-18\sqrt{3}\cdot i}=3\sqrt{3}-3i\) (and the non-principal root is \(-3\sqrt{3}+3i\))
2.4.3.8.
Answer.
\(\sqrt[3]{-16+16i}=2+2i\) (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.
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.
\(\sqrt[5]{-16\sqrt{3}-16i}=\sqrt{3}-i\) (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.
Answer.
\(\frac{3\sqrt{3}}{2}+\frac{3}{2}i\text{,}\) \(-3i\text{,}\) 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.
\(\left\{ -1, \frac{1}{2}+\frac{\sqrt{3}}{2}i, \frac{1}{2}-\frac{\sqrt{3}}{2}i \right\} \)
