Determining an Inverse Function (College Board AP® Precalculus): Study Guide

Range & domain of an inverse function

How are the domain and range of a function and its inverse related?

• When a function $f$ has an inverse $f^{-1}$, the inputs and outputs swap roles

• This means:

  • The domain of $f$ becomes the range of $f^{-1}$

  • The range of $f$ becomes the domain of $f^{-1}$

• E.g. if $f$ has domain $[0,5]$ and range $[2,10]$

  • then $f^{-1}$ has domain $[2,10]$ and range $[0,5]$

• This relationship also applies to values in tables

  • If you want to find the inverse by swapping input-output pairs

    • then the column of input values ($x$-values) in the original table becomes the column of output values in the inverse table

    • and vice versa

  • The point $(\mathit{a},\mathit{b})$ in $f$ corresponds to the point $(\mathit{b},\mathit{a})$ in $f^{-1}$

Are there additional restrictions on the applicability of an inverse function?

• Beyond the mathematical domain and range swap, contextual restrictions may also limit when an inverse function is meaningful

• E.g. if $f\left(t\right)$ models the height of a ball as a function of time

  • the inverse $f^{-1}\left(h\right)$ gives the time at which the ball reaches height $h$

    • But this inverse may only be useful for heights the ball actually reaches

    • and only for the time interval during which the ball is in the air

Finding an inverse function analytically

How do you find the inverse of a function from its formula?

• To find the inverse of a function given by an equation:

  • Start with the equation as $y=f\left(x\right)$

  • Then swap $x$ and $y$

    • this gives the equation as $x=f\left(y\right)$

  • Finally solve the resulting equation for $y$

    • The solution gives you $y=f^{-1}\left(x\right)$

• This method works by reversing the operations that $f$ performs on its input

• E.g. find the inverse of $f\left(x\right)=3x-5$:

  • Write

    • $y=3x-5$

  • Swap

    • $x=3y-5$

  • Solve

    • $x+5=3y$, so $y=\frac{x+5}{3}$

  • Therefore $f^{-1}\left(x\right)=\frac{x+5}{3}$

• E.g. find the inverse of $f\left(x\right)=2x^3-5$:

  • Write

    • $y=2x^3-5$

  • Swap

    • $x=2y^3-5$

  • Solve

    • $x+5=2y^3$, then $y^3=\frac{x+5}{2}$, so $y=\sqrt[3]{\frac{x+5}{2}}$

  • Therefore $f^{-1}\left(x\right)=\sqrt[3]{\frac{x+5}{2}}$

Composition of inverse functions

What happens when you compose a function with its inverse?

• As $f$ and $f^{-1}$ are inverses of each other, then composing them in either order returns the original input

  • $f\left(f^{-1}\right(x\left)\right)=x$

  • $f^{-1}\left(f\right(x\left)\right)=x$

• In other words, applying a function and then its inverse (or vice versa) "undoes" the effect

  • you end up back where you started

  • $f$ and $f^{-1}$ 'cancel each other'

• The composition of a function with its inverse (in either order) is equivalent to the identity function

  • I.e. the composition simply returns its input unchanged

• This property is useful for verifying a proposed inverse

  • Substitute one function into the other and check that the result simplifies to $x$

  • E.g. for $f\left(x\right)=3x-5$ and $f^{-1}\left(x\right)=\frac{x+5}{3}$:

    • $f\left(f^{-1}\right(x\left)\right)=3·\frac{x+5}{3}-5=(x+5)-5=x$ ✓

    • $f^{-1}\left(f\right(x\left)\right)=\frac{(3x-5)+5}{3}=\frac{3x}{3}=x$ ✓