Chapter 2.5 Practice
Questions
Question 1
- For each function \(f,\) determine whether it is invertible. If it is, identify its inverse.
- \(f=\{(1,1),(5,1),(6,2)\}\)
- \(f=\left\{\left(-\tfrac{1}{2},3\right),(-2,0),\left(2,\tfrac{3}{2}\right),\left(1,\tfrac{4}{3}\right)\right\}\)
- \(f=\left\{\left(\tfrac{5}{2},-\frac{4}{5}\right),(-1,1),\left(0,\tfrac{5}{2}\right)\right\}\)
Question 2
For each invertible function \(f\), determine the domain and range of \(f^{-1}\)
- the function \(f\) with \(\mathcal{D}(f)=(-\infty,\infty)\) and \(\mathcal{R}(f)=[3,\infty)\)
- the function \(f\) with \(\mathcal{D}(f)=(-\infty,5]\cup[6,\infty)\) and \(\mathcal{R}(f)=(-\infty,\infty)\)
- the function \(f\) with \(\mathcal{D}(f)=(2,7]\) and \(\mathcal{R}(f)=(-\infty,0]\cup(1,\infty)\)
- the function \(f\) whose graph is given below:
Question 3
Sketch each invertible function \(f\) and its inverse. Write a formula for its inverse.
- the function \(f|_A\) where \(f(x)=x^2\) and \(A=(-4,-3)\cup[0,3]\cup[4,5]\)
- the function \(f|_A\) where \(f(x)=x^2\) and \(A=(-6,-2)\cup[-1,0)\cup(1,2]\cup[6,7]\)
- the function \(f|_A\) where \(f(x)=(x-1)^2\) and \(A=(-3,-1]\cup[1,3)\)
- the function \(f\) whose graph is given below:
Question 4
- For each invertible function \(f\), determine \(f^{-1}\) as well as the domains and ranges of \(f\) and \(f^{-1}\)
- \(f(x)=3x+1\)
- \(f(x)=-\frac{2}{3}x-5\)
- \(f(x)=x^3+8\)
- \(f(x)=\frac{1}{2}(x-1)^3-2\)
- \(f(x)=\frac{3x}{10x+15}\)
- \(f(x)=\frac{6x-1}{4x+1}\)
Answers
Question 1
- Not invertible
- Invertible, \(f^{-1}=\left\{\left(3,-\tfrac{1}{2}\right),(0,-2),\left(\tfrac{3}{2},2\right),\left(\tfrac{4}{3},2\right)\right\}\)
- Invertible, \(f=\left\{\left(-\frac{4}{5},\tfrac{5}{2}\right),(1,-1),\left(\tfrac{5}{2},0\right)\right\}\)
Question 2
- \(\mathcal{D}(f^{-1})=[3,\infty)\), \(\mathcal{R}(f^{-1})=(-\infty,\infty)\)
- \(\mathcal{D}(f^{-1})=(-\infty,\infty)\), \(\mathcal{R}(f^{-1})=(-\infty,5]\cup[6,\infty)\)
- \(\mathcal{D}(f^{-1})=(-\infty,0]\cup(1,\infty)\), \(\mathcal{R}(f)=(2,7]\)
- \(\mathcal{D}(f^{-1})=[-4,0]\cup(4,8]\), \(\mathcal{R}(f^{-1})=[-3,-1)\cup[0,1]\)
Question 3
- The sketch of \(f|_A\) and its inverse is given below, while the formula is \[f^{-1}(x)=\begin{cases}\sqrt{x}&\text{if } 0\leq x\leq 9\\-\sqrt{x}&\text{if }9<x<16\\\sqrt{x}&\text{if }16\leq x\leq 25\end{cases}\]
- the function \(f|_A\) where \(f(x)=x^2\) and \(A=(-6,-2)\cup[-1,0)\cup(1,2]\cup[6,7]\) The sketch of \(f|_A\) and its inverse is given below, while the formula is \[f^{-1}(x)=\begin{cases}-\sqrt{x}&\text{if } 0< x\leq 1\\\sqrt{x}&\text{if }1<x\leq 4\\-\sqrt{x}&\text{if }4< x< 36\\\sqrt{x}&\text{if }36\leq x\leq 49\end{cases}\]
- The sketch of \(f|_A\) and its inverse is given below, while the formula is \[f^{-1}(x)=\begin{cases}\sqrt{x}+1&\text{if } 0\leq x< 4\\-\sqrt{x}+1&\text{if }4\leq x<16\end{cases}\]
- The sketch of \(f\) is already given. However, it may be helpful to write a formula for \(f\) to find its inverse: \[f(x)=\begin{cases}-2x+2&\text{if }-3\leq x<1\\4x-4&\text{if }0\leq 1\end{cases}.\] The sketch of its inverse is given below, while the formula is \[f^{-1}(x)=\begin{cases}-\frac{1}{2}(x-4)-1&\text{if } 4< x\leq 8\\\frac{1}{4}x+1&\text{if }-4\leq x<0\end{cases}\]
Question 4
- \(f^{-1}(x)=\frac{1}{3}x-\frac{1}{3}\);
\(\mathcal{D}(f)=(-\infty,\infty),\quad \mathcal{R}(f)=(-\infty,\infty)\)
\(\mathcal{D}(f^{-1})=(-\infty,\infty)\quad \mathcal{R}(f^{-1})=(-\infty,\infty)\) - \(f^{-1}(x)=-\frac{3}{2}x-\frac{15}{2}\);
\(\mathcal{D}(f)=(-\infty,\infty),\quad \mathcal{R}(f)=(-\infty,\infty)\),
\(\mathcal{D}(f^{-1})=(-\infty,\infty)\), \(\mathcal{R}(f^{-1})=(-\infty,\infty)\) - \(f^{-1}(x)=\sqrt[3]{x-8}\);
\(\mathcal{D}(f)=(-\infty,\infty), \quad \mathcal{R}(f)=(-\infty,\infty)\),
\(\mathcal{D}(f^{-1})=(-\infty,\infty),\quad \mathcal{R}(f^{-1})=(-\infty,\infty)\) - \(f^{-1}(x)=\sqrt[3]{2x+4}+1\);
\(\mathcal{D}(f)=(-\infty,\infty),\quad \mathcal{R}(f)=(-\infty,\infty)\),
\(\mathcal{D}(f^{-1})=(-\infty,\infty),\quad \mathcal{R}(f^{-1})=(-\infty,\infty)\) - \(f^{-1}(x)=-\frac{15x}{10x-3}\);
\(\mathcal{D}(f)=\left(-\infty,-\frac{3}{2}\right)\cup\left(-\frac{3}{2},\infty\right),\quad \mathcal{R}(f)=\left(-\infty,\frac{3}{10}\right)\cup\left(\frac{3}{10},\infty\right)\)
\(\mathcal{D}(f^{-1})=\left(-\infty,\frac{3}{10}\right)\cup\left(\frac{3}{10},\infty\right),\quad \mathcal{R}(f^{-1})=\left(-\infty,-\frac{3}{2}\right)\cup\left(-\frac{3}{2},\infty\right)\) - \(f^{-1}(x)=\frac{x+1}{-4x+6}\);
\(\mathcal{D}(f)=\left(-\infty,-\frac{1}{4}\right)\cup\left(-\frac{1}{4},\infty\right),\quad\mathcal{R}(f)=\left(-\infty,\frac{3}{2}\right)\cup\left(\frac{3}{2},\infty\right)\),
\(\mathcal{D}(f^{-1})=\left(-\infty,\frac{3}{2}\right)\cup\left(\frac{3}{2},\infty\right),\quad \mathcal{R}(f^{-1})=\left(-\infty,-\frac{1}{4}\right)\cup\left(-\frac{1}{4},\infty\right)\)
- \(f^{-1}(x)=\frac{1}{3}x-\frac{1}{3}\);