Inverse Functions (Edexcel IGCSE Maths A: Higher): Revision Note

Inverse functions

What is an inverse function?

• An inverse function does the opposite (reverse) operation of the function it came from

  • E.g. If a function “doubles the number then adds 1”

  • Then its inverse function “subtracts 1, then halves the result”

    • The same inverse operations are used when solving an equation or rearranging a formula

• An inverse function performs the inverse operations in the reverse order

What notation is used for inverse functions?

• The inverse function of $\mathrm{f}\left(x\right)$ is written as ${\mathrm{f}}^{-1}\left(x\right)=\dots$ or  ${\mathrm{f}}^{-1}:x\mapsto \dots$

  • For example, if $\mathrm{f}\left(x\right)=2x+1$

  • The inverse function is ${\mathrm{f}}^{-1}\left(x\right)=\frac{x-1}{2}$  or ${\mathrm{f}}^{-1}:x\mapsto \frac{x-1}{2}$

• If $\mathrm{f}\left(a\right)=b$ then ${\mathrm{f}}^{-1}\left(b\right)=a$

  • For example

    • $\mathrm{f}\left(3\right)=2\times 3+1=7$ (inputting 3 into $\mathrm{f}$ gives 7)

    • ${\mathrm{f}}^{-1}\left(7\right)=\frac{7-1}{2}=3$ (inputting 7 into ${\mathrm{f}}^{-1}$ gives back 3)

How do I find an inverse function algebraically?

• The process for finding an inverse function is as follows:

  • Write the function as $\mathit{y}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}$

    • E.g. The function $\mathrm{f}\left(x\right)=2x+1$ becomes $y=2x+1$

  • Swap the $x$s and $y$s to get $x=\dots$

    • E.g. $x=2y+1$

    • The letters change but no terms move

  • Rearrange the expression to make $\mathit{y}$ the subject again

    • E.g. $x=2y+1$ becomes $x-1=2y$ so $y=\frac{x-1}{2}$

  • Replace $\mathit{y}$ with ${\mathrm{f}}^{-1}\left(x\right)=\dots$(or ${\mathrm{f}}^{-1}:x\mapsto \dots$)

    • E.g. ${\mathrm{f}}^{-1}\left(x\right)=\frac{x-1}{2}$

    • This is the inverse function

    • $y$ should not appear in the final answer

How are inverse functions and composite functions related?

• The composite function of $\mathrm{f}$ followed by ${\mathrm{f}}^{-1}$ (or the other way round) cancels out

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

    • If you apply a function to x, then apply its inverse function, you get back x

    • Whatever happened to x gets undone

    • f and f^-1 cancel each other out when applied together

• For example, solve ${\mathrm{f}}^{-1}\left(x\right)=5$ where $\mathrm{f}\left(x\right)=2^x$

  • Finding the inverse function ${\mathrm{f}}^{-1}\left(x\right)$ algebraically in this case is tricky

    • (It is impossible if you haven't studied logarithms!)

  • Instead, you can take $\mathrm{f}$ of both sides of ${\mathrm{f}}^{-1}\left(x\right)=5$ and use the fact that ${\mathrm{ff}}^{-1}$ cancel each other out:
    

    • $\begin{array}{rcl}{\mathrm{ff}}^{-1}\left(x\right)& =& \mathrm{f}\left(5\right)\end{array}$ which cancels to $x=\mathrm{f}\left(5\right)$ giving $x=2^5=32$

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

• The domain of an inverse function has exactly the same values as the range of the original function

  • E.g. If $\mathrm{f}\left(x\right)=\frac{3}{x+1}$ has a range of $\mathrm{f}\left(x\right)>5$

    • then its inverse function, ${\mathrm{f}}^{-1}\left(x\right)=\frac{3}{x}-1$, has the domain $x>5$

    • Remember to always write domains in terms of $x$

• The range of an inverse function has exactly the same values as the domain of the original function

  • E.g. If $\mathrm{f}\left(x\right)=\frac{3}{x+1}$ has a domain of $x<-1$

    • then its inverse function, ${\mathrm{f}}^{-1}\left(x\right)=\frac{3}{x}-1$, has the range ${\mathrm{f}}^{-1}\left(x\right)<-1$

    • Remember to always write ranges in terms of their function, ${\mathrm{f}}^{-1}\left(x\right)$

How do I find the inverse of a quadratic function?

• You need to rewrite the quadratic expression by completing the square

  • E.g. if $\mathrm{f}\left(x\right)=x^2+4x-3$ then rewrite it as $\mathrm{f}\left(x\right)={\left(x+2\right)}^2-7$

• Follow the same process to find the inverse

  • E.g. $y={\left(x+2\right)}^2-7$

    • Swap: $x={\left(y+2\right)}^2-7$

    • Rearrange: $y=-2\pm \sqrt{x+7}$

• Use the domain of the quadratic function to decide whether to use the plus or minus root

  • The domain of the quadratic is the range of the inverse

  • E.g. suppose the domain of $\mathrm{f}\left(x\right)$ is $x\le -2$

    • Then ${\mathrm{f}}^{-1}\left(x\right)=-2-\sqrt{x+7}$

  • E.g. suppose the domain $\mathrm{f}\left(x\right)$ is $x\ge -2$

    • Then ${\mathrm{f}}^{-1}\left(x\right)=-2+\sqrt{x+7}$