Inverse Functions (Cambridge International IGCSE Mathematics)

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 equation$\mathrm{f} ( x )$ is written as $f^{-1}(x)=\dots$

  • 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 $x$)

$-16\leq x<11$