1.
\(f^{-1}(x) =\)
Hint.
Example: Find the inverse function of \(f(x) = \sqrt{x+1}\)
\(\displaystyle{\begin{aligned}
y \amp = \sqrt{x+1}\\
y^2 \amp = x+1\\
y^2-1 \amp =x\\
\end{aligned}
}\)
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Worksheet Inverse Functions
2.
\(f^{-1}(x) =\)
Hint.
Example: Find the inverse function of \(f(x) = \frac{2}{x-3} +4\)
\(\displaystyle{\begin{aligned}
y \amp = \frac{2}{x-3} +4\\
y-4\amp =\frac{2}{x-3}\\
(y-4)(x-3) \amp =2\\
x-3 \amp =\frac{2}{y-4}\\
x\amp =\frac{2}{y-4}+3\\
\end{aligned}
}\)
Therefore, \(\displaystyle f^{-1}(y) = \frac{2}{y-4}+3\)
or, \(\displaystyle f^{-1}(x) = \frac{2}{x-4}+3\)
3.
\(f^{-1}(x) =\)
4.
5.
Let the graph of \(f(x)\) be given below. Find the following.
(Click on graph to enlarge)
\(f(0) =\)
\(f^{-1}(0) =\)
6.
(a) Let the graph of \(f(x)\) be given by:
(Click on graph to enlarge)
Which of the following is the graph of \(y=f^{-1}(x)\text{?}\)
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A
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B
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C
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D
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| A | B |
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|
| C | D |
(Click on a graph to enlarge it.)
(b) Let the graph of \(g(x)\) be given by:
(Click on graph to enlarge)
Which of the following is the graph of \(y=g^{-1}(x)\text{?}\)
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A
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B
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C
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D
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| A | B |
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|
| C | D |
(Click on a graph to enlarge it.)
7.
Use the table to evaluate the expressions:
\(f(2)=\)
\(f(9)=\)
\(f^{-1}(-6)=\)
\(f^{-1}(-13)=\)
8.
An object dropped from a height of \(400\) feet has a height, \(h(t)\text{,}\) in feet after \(t\) seconds have elapsed, where \(t\ge 0\text{,}\) such that \(h(t)={400-16t^{2}}\text{.}\) Express \(t\) as a function of height \(h\text{,}\) and find the time to reach a height of \(150\) feet.
9.
Let \(\displaystyle{f(x) = {\left(x-3\right)^{2}}}\text{.}\)
Find a restricted domain in which the function \(f(x)\) is \(one\,to\,one\text{,}\) non-decreasing, and on which \(f(x)\) achieves its entire range. Use interval notation.
Find \(f^{-1}(x)\) on the restricted domain.
\(f^{-1}(x) =\)
10.
Determine if the following functions are \(one\,to\,one\text{.}\) If the function is one to one then the function has an inverse.
Hint: Use the horizontal line test.
11.
12.
13.
Are the following functions one-to-one?
Solution.
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\(\begin{aligned} y\amp ={x^{3}+5}\\ y-5\amp =x^3\\ x^3\amp =y-5\\ x\amp =\sqrt[3]{y-5}\\ \end{aligned}\)Note that in the last step, it is legitimate to take a 3rd root of both sides of the equation because 3 is odd. Since we can solve for \(x\) in terms of \(y\text{,}\) \({x^{3}+5}\) is a one-to-one function of \(x\text{.}\)
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\(\begin{aligned} y\amp ={\left|x-8\right|}\\ {\left|x-8\right|}\amp =y\\ \end{aligned}\)At this point we cannot determine what \(x\) is exactly. Maybe \(x-8=y\) or maybe \(x-8=-y\text{.}\) So maybe \(x=y+8\) or maybe \(x=-y+8\text{.}\) Since we cannot solve for \(x\) in terms of \(y\text{,}\) \({\left|x-8\right|}\) is not a one-to-one function of \(x\text{.}\)
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\(\begin{aligned} y\amp ={5x+4}\\ {5x+4}\amp =y\\ 5x\amp =y-4\\ x\amp =\frac{y-4}{5}\\ \end{aligned}\)Since we can solve for \(x\) in terms of \(y\text{,}\) \({5x+4}\) is a one-to-one function of \(x\text{.}\)
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\(\begin{aligned} y\amp ={\frac{1}{x}}\\ xy\amp =1\\ x\amp =\frac{1}{y}\\ \end{aligned}\)Since we can solve for \(x\) in terms of \(y\text{,}\) \({\frac{1}{x}}\) is a one-to-one function of \(x\text{.}\)
14.
(a) Find \(f(4)\text{.}\)
(b) Find \(g(2)\text{.}\)
(c) Find \((g\circ f)(1)\text{.}\)
(d) Find \(g^{-1}(5)\text{.}\)
15.
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\(\displaystyle f^{-1}(x) = \displaystyle \frac{3x-2}{6x-3}\)
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\(\displaystyle f^{-1}(x) = \displaystyle \frac{3x-4}{5x+4}\)
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\(\displaystyle f^{-1}(x) = \displaystyle \frac{4-2x}{4-4x}\)
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\(\displaystyle f^{-1}(x) = \displaystyle \frac{5x+6}{4x-5}\)
16.
17.
Consider the function
\begin{equation*}
f(x) = \frac{x}{2x-4}.
\end{equation*}
a) Find the inverse function for \(f\)
\(f^{-1}(x)\)= ,
18.
Find the inverse function of \(g(x) = \frac{ \sqrt{x} + 2}{7 - \sqrt{x}}\text{.}\) If the function is not invertible, enter NONE.
\(g^{-1}(x) =\)
(Write your inverse function in terms of the independent variable \(x\text{.}\))
Solution.
SOLUTION\(g \left( g^{-1}(x) \right) = x\text{,}\)\(y\)\(g^{-1}(x)\)\(g(y)=x\text{.}\)\(g\)\(\displaystyle g(y) = \frac{ \sqrt{y} + 2}{7 - \sqrt{y}} =x\)\(y\)
\begin{equation*}
\begin{aligned}
x \amp = g(y) = \frac{ \sqrt{y} + 2}{7 - \sqrt{y}} \\
x(7 - \sqrt{y}) \amp = \sqrt{y}+ 2 \\
7 x-x \sqrt{y} \amp = \sqrt{y} + 2 \\
7 x - 2 \amp = \sqrt{y}+x\sqrt{y} \\
7 x- 2 \amp = \sqrt{y}(1+x) \\
\sqrt{y} \amp = \frac{7 x - 2}{1+x} \\
y \amp = \left( \frac{7 x - 2}{1+x} \right)^2.
\end{aligned}
\end{equation*}
\(f(x)\)\((-\infty,-1)\cup[2/7,\infty)\text{,}\)
\begin{equation*}
f^{-1}(x)=\left(\frac{7 x-2}{1+x}\right)^2, \quad x\in
(-\infty,-1)\cup[2/7,\infty).
\end{equation*}
