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As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function , particularly of rational functions that are the ratio of linear functions, such as in concentration applications.
The function $\text{\hspace{0.17em}}C=\frac{20+0.4n}{100+n}\text{\hspace{0.17em}}$ represents the concentration $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ of an acid solution after $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ mL of 40% solution has been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}C.\text{\hspace{0.17em}}$ Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution.
We first want the inverse of the function in order to determine how many mL we need for a given concentration. We will solve for $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}C.$
Now evaluate this function at 35%, which is $\text{\hspace{0.17em}}C=\mathrm{0.35.}$
We can conclude that 300 mL of the 40% solution should be added.
Find the inverse of the function $\text{\hspace{0.17em}}f(x)=\frac{x+3}{x-2}.$
${f}^{-1}(x)=\frac{2x+3}{x-1}$
Access these online resources for additional instruction and practice with inverses and radical functions.
Explain why we cannot find inverse functions for all polynomial functions.
It can be too difficult or impossible to solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}y.$
Why must we restrict the domain of a quadratic function when finding its inverse?
When finding the inverse of a radical function, what restriction will we need to make?
We will need a restriction on the domain of the answer.
The inverse of a quadratic function will always take what form?
For the following exercises, find the inverse of the function on the given domain.
$f\left(x\right)={\left(x-4\right)}^{2},[4,\infty )$
$\text{\hspace{0.17em}}\text{\hspace{0.17em}}{f}^{-1}(x)=\sqrt{x}+4$
$f\left(x\right)={\left(x+2\right)}^{2},[\mathrm{-2},\infty )$
$f(x)={\left(x+1\right)}^{2}-3,[\mathrm{-1},\infty )$
$\text{\hspace{0.17em}}\text{\hspace{0.17em}}{f}^{-1}(x)=\sqrt{x+3}-1$
$f(x)=3{x}^{2}+5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\infty ,0\right]$
${f}^{-1}(x)=-\sqrt{\frac{x-5}{3}}$
$f\left(x\right)=12-{x}^{2},[0,\infty )$
$f(x)=2{x}^{2}+4,[0,\infty )$
For the following exercises, find the inverse of the functions.
$f(x)={x}^{3}+5$
$\text{\hspace{0.17em}}\text{\hspace{0.17em}}{f}^{\mathrm{-1}}(x)=\sqrt[3]{x-5}$
$f\left(x\right)=3{x}^{3}+1$
$f(x)=4-{x}^{3}$
$\text{\hspace{0.17em}}{f}^{-1}(x)=\sqrt[3]{4-x}$
$f\left(x\right)=4-2{x}^{3}$
For the following exercises, find the inverse of the functions.
$f(x)=\sqrt{2x+1}$
${f}^{\mathrm{-1}}(x)=\frac{{x}^{2}-1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[0,\infty \right)$
$f(x)=\sqrt{3-4x}$
$f\left(x\right)=9+\sqrt{4x-4}$
$\text{\hspace{0.17em}}{f}^{\mathrm{-1}}(x)=\frac{{\left(x-9\right)}^{2}+4}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[9,\infty \right)$
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