Composite & Inverse Functions (AQA GCSE Further Maths): Revision Note

Composite functions

What is a composite function?

• A composite function is one function applied to the output of another function

• Composite functions may also be referred to as compound functions

What do composite functions look like?

• The notation you will see for a composite function is fg(x)

  • This can be written as f(g(x)) and means “f applied to the output of g(x)” 

  • i.e. g(x) happens first

• Always apply the function on the outside to the output of the function on the inside

  • gf(x) means g(f(x)) and means “g applied to the output of f(x)” 

  • i.e. f(x) happens first

How does a composite function work?

• If you are putting a number into fg(x)

  • STEP 1
    Put the number into g(x)

  • STEP 2
    Put the output of g(x) into f(x)

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

    • $\mathrm{fg}\left(2\right)=\mathrm{f}\left(\frac{1}{2}\right)=2\times \frac{1}{2}+1=2$

    • $\mathrm{gf}\left(2\right)=\mathrm{g}(2\times 2+1)=\mathrm{g}\left(5\right)=\frac{1}{5}$

• If you are using algebra, to find an expression for a composite function

  • STEP 1
    For fg(x) put g(x) wherever you see x in f(x)

  • STEP 2
    Simplify if necessary

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

    • $\mathrm{fg}\left(x\right)=\mathrm{f}\left(\frac{1}{x}\right)=2\times \frac{1}{x}+1=\frac{2}{x}+1$

    • $\mathrm{gf}\left(x\right)=\mathrm{g}(2x+1)=\frac{1}{2x+1}$

Inverse functions

What is an inverse function?

• An inverse function does the exact opposite of the function it came from

  • For example, if the function “doubles the number and adds 1” then its inverse is

  • “subtract 1 and halve the result”

• It is the inverse operations in the reverse order

How do I write inverse functions?

• An inverse function f^-1 can be written as ${\mathrm{f}}^{-1}\left(x\right)=\dots$ 

  • For example, if $\mathrm{f}\left(x\right)=2x+1$ its inverse can be written as

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

How do I find an inverse function?

• The easiest way to find an inverse function is to 'cheat' and swap the $x$ and $y$ variables

  • Note that this is a useful method but you MUST remember not to do this in any other circumstances in maths

  • STEP 1
    Write the function in the form$y=\dots$

  • STEP 2
    Swap the $x$s and $y$s to get $x=\dots$

  • STEP 3
    Rearrange the expression to make $y$ the subject again

  • STEP 4
    Write as f^-1(x) = … 
    

    • $y$ should not exist in the final answer

• For example, if $\mathrm{f}\left(x\right)=2x+1$ its inverse can be found as follows 

  • STEP 1
    Write the function in the form $y=2x+1$

  • STEP 2
    Swap the $x$ and $y$ to get $x=2y+1$

  • STEP 3
    Rearrange the expression to make $y$ the subject again

$\begin{array}{rcl}x-1& =& 2y\\ \frac{x-1}{2}& =& y\to y=\frac{x-1}{2}\end{array}$

• STEP 4
  Rewrite using the correct notation for an inverse function
  

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

How does a function relate to its inverse?

• If $\mathrm{f}\left(3\right)=10$ then the input of 3 gives an output of 10

  • The inverse function undoes f(x)

  • An input of 10 into the inverse function gives an output of 3

    • If $f\left(3\right)=10$ then ${\mathrm{f}}^{-1}\left(10\right)=3$

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

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

    • Whatever happened to x gets undone

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

• If $\mathrm{f}\left(x\right)=2^x$ and you want to solve ${\mathrm{f}}^{-1}\left(x\right)=5$

  • Finding the inverse function ${\mathrm{f}}^{-1}\left(x\right)$ in this case is tricky (impossible if you haven't studied logarithms)

  • instead, take f of both sides and use that ${\mathrm{ff}}^{-1}$ cancel each other out:

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