Practice 1 Solutions

1. Explain how the inverse of a function is found algebraically.

The inverse of a function is found by switching the domain and range values, or in other words switching the variables x and y in the equation.

2. $R=\left\{\left(5, 2\right), \left(6, –7\right), \left(3, 1\right), \left(8, –10\right)\right\}$

The relation and its inverse are both functions.

3. $R=\left\{\left(–3, 4\right), \left(9, –2\right), \left(6, 1\right), \left(–4, 3\right), \left(2, 1\right)\right\}$

The relation is a function, but the inverse is not a function.

4. A class was asked to list their favorite summer activity. The results were:
   R = {(Maddie, hiking), (Hope, relaxing), (Stetson, swimming), (Austin, hiking), (Natalee, relaxing)}.

The relation is a function, but the inverse is not a function.

5. $f\left(x\right)=9{\left(x–2\right)}^2–1 g\left(x\right)=\frac{1}{3}\sqrt{x+1}+2$

$\begin{array}{ccc}f\left(4\right)=9{\left(4–2\right)}^2–1& & g\left(35\right)=\frac{1}{3}\sqrt{35+1}+2\\ f\left(4\right)=35& & g\left(35\right)=4 \end{array}$

$f\left(x\right)$ and $g\left(x\right)$ are inverses

6. $f\left(x\right)=\frac{5}{4}x–2 g\left(x\right)=\frac{4}{5}x+2$

$\begin{array}{ccc} f\left(4\right)=\frac{5}{4}\left(4\right)–2& & g\left(3\right)=\frac{4}{5}\left(3\right)+2\\ f\left(4\right)=3& & g\left(3\right)=\frac{22}{5}\end{array}$

$f\left(x\right)$ and $g\left(x\right)$ are not inverses

7. $f\left(x\right)=2\sqrt[3]{x–4}+5 g\left(x\right)=\frac{1}{2}{\left(x+5\right)}^3+4$

$\begin{array}{ccc}f\left(4\right)=2\sqrt[3]{4–4}+5& & g\left(5\right)=\frac{1}{2}{\left(5+5\right)}^3+4\\ f\left(4\right)=5& & g\left(5\right)=7816.5\end{array}$

$f\left(x\right)$ and $g\left(x\right)$ are not inverses

8. $f\left(x\right)=7x+9$, check with $x=6.$

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{ccc}x=7x+9 \\ x–9=7y \\ y=\frac{x–9}{7}& & f\left(6\right)=7\left(6\right)+9 \\ f\left(6\right)=51\\ & & f^{–1}\left(51\right)=\frac{51–9}{7} \\ {f}^{–1}\left(51\right)=6\end{array}\end{array} \end{gathered}$$

$f^{–1}\left(x\right)=\frac{x–9}{7}$

9. $f\left(x\right)=2\sqrt{x}–3 \mathrm{for} \left\{x|x\ge 0\right\}$, check with $x=9$.

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{ccc}x=2\sqrt{y}–3 \\ x+3=2\sqrt{y} \\ \frac{x+3}{2}=\sqrt{y} \\ y={\left(\frac{x+3}{2}\right)}^2& & f\left(9\right)=2\sqrt{9}–3 \\ f\left(9\right)=3\\ & & f^{–1}\left(3\right)=\frac{1}{4}{\left(3+3\right)}^2 \\ {f}^{–1}\left(3\right)=9\end{array}\end{array} \end{gathered}$$

$f^{–1}\left(x\right)=\frac{1}{4}{\left(x+3\right)}^2$

10. $f\left(x\right)=8{\left(x+1\right)}^3$, check with $x=– 2.$

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{ccc}x=8{\left(y+1\right)}^3 \\ \frac{x}{8}={\left(y+1\right)}^3 \\ \sqrt[3]{\frac{x}{8}}=y+1 \\ \frac{1}{2}\sqrt[3]{x}=y+1 \\ y=\frac{1}{2}\sqrt[3]{x}–1& & f\left(–2\right)=8{\left(–2+1\right)}^3 \\ f\left(–2\right)=–8\\ & & f^{–1}\left(–8\right)=\frac{1}{2}\sqrt[3]{–8}–1\\ & & f^{–1}\left(–8\right)=–2\end{array}\end{array} \end{gathered}$$

$\begin{array}{rcl} f^{–1}\left(x\right)& =& \frac{1}{2}\sqrt[3]{x}–1\end{array}$

11. $f\left(x\right)=\frac{1}{x–4}+1$, check with $x=8.$

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{ccc}x=\frac{1}{y–4}+1 \\ x–1=\frac{1}{y–4} \\ \left(x–1\right)\left(y–4\right)=1 \\ y–4=\frac{1}{x–1}& & f\left(8\right)=\frac{1}{8–1}+4 \\ f\left(8\right)=1\frac{1}{4}\\ & & f^{–1}\left(1\frac{1}{4}\right)=\frac{1}{1{\displaystyle \frac{1}{4}}–1}+4\\ & & f^{–1}\left(1\frac{1}{4}\right)=8\end{array}\end{array} \end{gathered}$$

$\begin{array}{rcl} f^{–1}\left(x\right)& =& \frac{1}{x–1}+4\\ & & {\mathrm{Domain}}_f: \left\{x\in \mathrm{ℝ}, x\ne 4\right\}, {\mathrm{Range}}_f: \left\{y\in \mathrm{ℝ}, y\ne 1\right\}\\ & & {\mathrm{Domain}}_{f^{–1}}: \left\{x\in \mathrm{ℝ}, x\ne 1\right\}, {\mathrm{Range}}_{f^{–1}}: \left\{y\in \mathrm{ℝ}, y\ne 4\right\}\end{array}$

12. $f\left(x\right)={\left(x+1\right)}^2+3, \left\{x|x\ge –1\right\}$, check with $x=5.$

$$\begin{gathered} \begin{array}{rcl}& & \begin{array}{ccc}x={\left(y+1\right)}^2+3 \\ x–3={\left(y+1\right)}^2 \\ \pm \sqrt{x–3}=y+1& & f\left(5\right)={\left(5+1\right)}^2+3 \\ f\left(5\right)=39\\ & & f^{–1}\left(39\right)=\sqrt{39–3}–1 \\ {f}^{–1}\left(39\right)=5\end{array}\end{array} \end{gathered}$$

$\begin{array}{rcl} f^{–1}\left(x\right)& =& \sqrt{x–3}–1\\ & & {\mathrm{Domain}}_f: \left\{x\in \mathrm{ℝ}, x\ge –1\right\}, {\mathrm{Range}}_f: \left\{y\in \mathrm{ℝ}, y\ge 3\right\}\\ & & {\mathrm{Domain}}_{f^{–1}}: \left\{x\in \mathrm{ℝ}, x\ge 3\right\}, {\mathrm{Range}}_{f^{–1}}: \left\{y\in \mathrm{ℝ}, y\ge –1\right\}\end{array}$