Inverse Functions (DP IB Analysis & Approaches (AA): HL): Revision Note

Inverse functions

What is an inverse function?

• An inverse function, $f^{-1}\left(x\right)$, reverses (or undoes) the effect of $f\left(x\right)$

  • for example

    • if $f\left(x\right)=2x$ then $f^{-1}\left(x\right)=\frac{x}{2}$

    • if $g\left(x\right)=x+10$ then $g^{-1}\left(x\right)=x-10$

• Inverse functions can be used to solve equations

  • e.g. the solution of $f\left(x\right)=8$ is $x=f^{-1}\left(8\right)$

What is the identity function?

• The identity function $\mathrm{id}$ maps each value to itself

  • $\mathrm{id}\left(x\right)=x$

    • e.g. $\mathrm{id}\left(5\right)=5$

• Applying a function $f$ to an input

  • then applying the inverse, $f^{-1}$ 

    • gives back the original input

• This also works if you swap the order of $f$ and $f^{-1}$ 

  • i.e. the composite function $(f\circ f^{-1})\left(x\right)=(f^{-1}\circ f)\left(x\right)=x$

    • has the same effect as the identity function

How do I sketch an inverse function?

• The graph of $y=f^{-1}\left(x\right)$ is a reflection of the graph $y=f\left(x\right)$ in the line $y=x$

  • e.g. if $f\left(x\right)=2x$ and $f^{-1}\left(x\right)=\frac{x}{2}$

    • then $y=2x$ and $y=\frac{1}{2}x$

    • which reflect in $y=x$

• If $y=f\left(x\right)$ intersects $y=x$ then $f^{-1}\left(x\right)$ also intersects $y=x$ at the same point

  • i.e. solutions to either $f\left(x\right)=x$ or $f^{-1}\left(x\right)=x$

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

  • There may be other solutions to $f\left(x\right)=f^{-1}\left(x\right)$ that don't lie on the line $y=x$

How do I find the inverse of a function?

• To find the inverse function using algebra, following these steps:

• STEP 1
  Swap the $x$ and $y$ in $y=f\left(x\right)$

• STEP 2
  Rearrange $x=f\left(y\right)$ to make $y$ the subject

  • The result is $f^{-1}\left(x\right)$

How do I find the domain and range of an inverse function?

• The domain of a function becomes the range of its inverse

  • e.g. if $f\left(x\right)=2x$ has domain $1\le x\le 3$

    • then the range of $f^{-1}\left(x\right)$ is $1\le f^{-1}\left(x\right)\le 3$

• The range of a function becomes the domain of its inverse

  • e.g. if $f\left(x\right)=2x$ has range $2\le f\left(x\right)\le 6$

    • then the domain of $f^{-1}\left(x\right)$ is $2\le x\le 6$

What condition is needed for an inverse function to exist?

• For an inverse function $f^{-1}\left(x\right)$ to exist

  • the original function $f\left(x\right)$ must be one-to-one

• This ensures $f^{-1}\left(x\right)$ never gives out two or more outputs

  • which functions are not allowed to do

How do I restrict many-to-one functions to be one-to-one?

• To restrict the domain of a many-to-one function

  • choose a subset of the domain on which the function is one-to-one

    • e.g. restrict $x\in \mathrm{ℝ}$ for the function $f\left(x\right)=x^2$ to $x\ge 0$

    • or $x\le 0$ or $1\le x\le 2$ or ... etc

• To find the biggest possible one-to-one domain:

• For quadratics

  • use the vertex as the upper or lower bound for the restricted domain

  • e.g. $f\left(x\right)=x^2$

    • $x\ge 0$ or $x\le 0$

  • e.g. $f\left(x\right)=a{(x-h)}^2+k$

    • $x\ge h$ or $x\le h$

• For trigonometric functions

  • use part of a cycle

  • e.g. $f\left(x\right)=\sin x$

    • $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

  • e.g. $f\left(x\right)=\cos x$

    • $0\le x\le \pi$

  • e.g. $f\left(x\right)=\tan x$ restrict the domain to one cycle between two asymptotes

    • $-\frac{\pi }{2}<x<\frac{\pi }{2}$

How do I find the inverse function of a restricted function?

• This is best shown through an example

• e.g. to find the inverse of the function $f\left(x\right)=x^2$ that has been restricted to the domain $x\ge 0$

  • use algebra initially

    • $x=y^2$ gives $y=\pm \sqrt{x}$

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

  • then choose a sign depending on the range of the inverse

    • Use the rule that "the range of the inverse, $f^{-1}\left(x\right)$ is equals the domain of the function"

    • The domain of $f$ is $x\ge 0$

    • so $f^{-1}\left(x\right)\ge 0$ is the range of $f^{-1}$

  • $f^{-1}\left(x\right)\ge 0$ means the inverse function always gives positive outputs

    • so choose the positive square root out of $f^{-1}\left(x\right)=\pm \sqrt{x}$

    • The answer is $f^{-1}\left(x\right)=\sqrt{x}$

• If the restricted domain is changed to $x\le 0$

  • the inverse function changes to$f^{-1}\left(x\right)=-\sqrt{x}$