Inverse Functions (Edexcel IGCSE Maths)

Spec point — 3.3I

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 $\mathrm{f} ( x )$ is written as $f^{-1}(x)=\dots$ or f ^ - 1 colon x ->from bar …

  • For example, if $f(x)=2x+1$
  • The inverse function is $f^{-1}(x)=\frac{x-1}{2}$ or $\mathrm{f}^{- 1} : x \rightarrow \text{from} \text{bar} \frac{x - 1}{2}$

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

  • For example

    • $\mathrm{f} ( 3 ) = 2 \times 3 + 1 = 7$ (inputting 3 into $f$ gives 7)
    • $\mathrm{f}^{- 1} ( 7 ) = \frac{7 - 1}{2} = 3$ (inputting 7 into $f^{-1}$ 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 $y=...$

    • E.g. The function $f(x)=2x+1$ 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$y$the subject again

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

  • Replace $y$ with $f^{-1}(x)=\dots$(or f ^ - 1 colon x ->from bar …)

    • E.g. $\mathrm{f}^{- 1} ( x ) = \frac{x - 1}{2}$
    • 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 $f$ followed by $f^{-1}$ (or the other way round) cancels out

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

    • 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^-1cancel each other out when applied together

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

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

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

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

    • $\begin{gathered} ff^{- 1} ( x ) & = \mathrm{f} ( 5 ) \end{gathered}$ which cancels to $x = \mathrm{f} ( 5 )$ 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 $\mathrm{f} ( x ) = \frac{3}{x + 1}$ has a range of $\mathrm{f} ( x ) > 5$

    • then its inverse function, $\mathrm{f}^{- 1} ( x ) = \frac{3}{x} - 1$, 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 $\mathrm{f} ( x ) = \frac{3}{x + 1}$ has a domain of $x<-1$

    • then its inverse function, $\mathrm{f}^{- 1} ( x ) = \frac{3}{x} - 1$, has the range $\mathrm{f}^{- 1} ( x ) < - 1$
    • Remember to always write ranges in terms of their function, $\mathrm{f}^{- 1} ( x )$

Worked Example A function $\mathrm{f} ( x ) = 5 - 3 x$has the domain $-2<x\leq7$.

(a) Use algebra to find $\mathrm{f}^{- 1} ( x )$.

Answer:

Write the function in the form $y=5-3x$ and then swap the $x$ and $y$

$y=5-3x\,x=5-3y$

*Rearrange **the expression to make *$y$ the subject again

$\begin{gathered} x = 5 - 3 y \\ x + 3 y & = 5 \\ 3 y & = 5 - x \\ y = \frac{5 - x}{3} \end{gathered}$

Rewrite the answer using inverse function notation

$f^{-1}(x)=\frac{5-x}{3}$

(b) Find the domain of $\mathrm{f}^{- 1} ( x )$.

Answer:

The domain of the inverse function is the range of the original function

Find the range of $\mathrm{f} ( x )$ by first finding $\mathrm{f} ( - 2 )$ and $\mathrm{f} ( 7 )$

$\begin{gathered} \mathrm{f} ( - 2 ) & = 5 - 3 ( - 2 ) = 5 + 6 = 11 \\ \mathrm{f} ( 7 ) & = 5 - 3 ( 7 ) = 5 - 21 = - 16 \end{gathered}$

The graph of $y=5-3x$ is a straight line with a negative gradient Between x = -2 and x = 7 the graph decreases from a height of 11 to a height of -16

The range of $\mathrm{f} ( x )$ is $- 16 \le \mathrm{f} ( x ) < 11$

Note that the inequality is "equal to" at x = 7, f(x) = -16 (this is the opposite order of "equal to" in the domain)

The domain of $\mathrm{f}^{- 1} ( x )$ takes the same values as range of $\mathrm{f} ( x )$ Write down the domain of $\mathrm{f}^{- 1} ( x )$* * (Remember that domains are always written in terms of $x$)

$-16\leq x<11$

How do I find the inverse of a quadratic function?

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

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

• Follow the same process to find the inverse

  • E.g. $y = ( x + 2 )^{2} - 7$

    • Swap: $x = ( y + 2 )^{2} - 7$
    • 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 $\mathrm{f} ( x )$ is $x\leq-2$

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

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

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

Worked Example Let $\mathrm{f} : x \rightarrow \text{from} \text{bar} 2 x^{2} - 12 x + 1$ for $x\geq3$

(a) Express the inverse function $f^{-1}$ in the form $\mathrm{f}^{- 1} : x \rightarrow \text{from} bar...$

Answer:

Complete the square for the quadratic expression

$2 ( x^{2} - 6 x ) + 1 2 [ ( x - 3 )^{2} - 9 ] + 1 2 ( x - 3 )^{2} - 18 + 1 2 ( x - 3 )^{2} - 17$

Set the function equal to $y$ and then swap $x$ and $y$

$y = 2 ( x - 3 )^{2} - 17 x = 2 ( y - 3 )^{2} - 17$

Rearrange to make $y$ the subject

$\begin{gathered} ( y - 3 )^{2} & = \frac{x + 17}{2} \\ y - 3 & = \pm \sqrt{ \frac{x + 17}{2} } \\ y = 3 \pm \sqrt{ \frac{x + 17}{2} } \end{gathered}$

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

• Therefore $y\geq3$

• Take the positive root

Write in the required format

$\mathrm{f}^{- 1} : x \rightarrow \text{from} \text{bar} 3 + \sqrt{ \frac{x + 17}{2} }$

(b) State the domain of $f^{-1}$

Answer:

Find the range of $f$

When $x\geq3$, $( x - 3 ) ^{2} \ge 0$ so $2 ( x - 3 ) ^{2} - 17 \ge - 17$

$\mathrm{f} ( x ) \ge - 17$

The range of a function is the domain of its inverse

• Write the domain in terms of $x$

$x\geq-17$