<< Chapter < Page | Chapter >> Page > |
Draw graphs of the functions $\text{\hspace{0.17em}}f\text{}$ and $\text{}{f}^{-1}$ from [link] .
Is there any function that is equal to its own inverse?
Yes. If $\text{\hspace{0.17em}}f={f}^{-1},\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}f\left(f\left(x\right)\right)=x,\text{\hspace{0.17em}}$ and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because
Any function $\text{\hspace{0.17em}}f\left(x\right)=c-x,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a constant, is also equal to its own inverse.
Access these online resources for additional instruction and practice with inverse functions.
Visit this website for additional practice questions from Learningpod.
Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?
Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that $\text{\hspace{0.17em}}y$ -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no $\text{\hspace{0.17em}}y$ -values repeat and the function is one-to-one.
Why do we restrict the domain of the function $\text{\hspace{0.17em}}f(x)={x}^{2}\text{\hspace{0.17em}}$ to find the function’s inverse?
Can a function be its own inverse? Explain.
Yes. For example, $\text{\hspace{0.17em}}f(x)=\frac{1}{x}\text{\hspace{0.17em}}$ is its own inverse.
Are one-to-one functions either always increasing or always decreasing? Why or why not?
How do you find the inverse of a function algebraically?
Given a function $\text{\hspace{0.17em}}y=f(x),\text{\hspace{0.17em}}$ solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Interchange the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Solve the new equation for $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ The expression for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the inverse, $\text{\hspace{0.17em}}y={f}^{-1}(x).$
Show that the function $\text{\hspace{0.17em}}f(x)=a-x\text{\hspace{0.17em}}$ is its own inverse for all real numbers $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$
For the following exercises, find $\text{\hspace{0.17em}}{f}^{-1}(x)\text{\hspace{0.17em}}$ for each function.
$f(x)=x+5$
$f(x)=3-x$
$f(x)=\frac{2x+3}{5x+4}$
For the following exercises, find a domain on which each function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ restricted to that domain.
$f(x)={(x+7)}^{2}$
domain of $f(x):\text{\hspace{0.17em}}[-7,\infty );\text{\hspace{0.17em}}{f}^{-1}(x)=\sqrt{x}-7$
$f(x)={(x-6)}^{2}$
$f(x)={x}^{2}-5$
domain of $\text{\hspace{0.17em}}f(x):\text{\hspace{0.17em}}[0,\infty );\text{\hspace{0.17em}}{f}^{-1}(x)=\sqrt{x+5}$
Given $\text{\hspace{0.17em}}f\left(x\right)=\frac{x}{2+x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g(x)=\frac{2x}{1-x}:$
a. $\text{\hspace{0.17em}}f(g(x))=x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g(f(x))=x.\text{\hspace{0.17em}}$ b. This tells us that $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are inverse functions
Notification Switch
Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?