# Composite and inverse functions

### 1) Composite functions

Two functions can be composed to form a new composite function. To write the composition of functions $f(x)$ and $g(x)$, write $f(x) \circ g(x)$ or $f(g(x))$. A composite function is sometimes called a compound function.

Question 1

Find the composition of the two functions:
$f(x)=x^{2}-2x+3$ and $g(x)=2x-1$.

$(f \circ g)(x)=(2x-1)^{2}-2(2x-1)+3=4x^{2}-8x+6$
$(g \circ f)(x)=2(x^{2}-2x+3)-1=2x^{2}-4x+5$.

Notice that $f\circ g \neq g\circ f$.

If you’re feeling confident, try this more difficult example. Don’t be worried if it looks hard, the process you go through is just the same as before. Replace with the function and then simplify.

Consider the functions $p(x)=\sqrt{x-1}$ and $q(x)=x^{2}$. Find $(q \circ p)(x)$.

Question 2

$f(x)=\frac{x-1}{x+1}$, find $h(x)=f(f(x))$
$f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}$

You now need to simplify the numerator and denominator separately.

Numerator  $\frac{x-1}{x+1}-1=\frac{x-1}{x+1}-\frac{x+1}{x+1}=\frac{x-1-(x+1)}{x+1}=\frac{-2}{x+1}$
Denominator  $\frac{x-1}{x+1}+1=\frac{x-1}{x+1}+\frac{x+1}{x+1}=\frac{x-1+(x+1)}{x+1}=\frac{2x}{x+1}$
$f(f(x))=\frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}=\frac{-2}{x+1}\times\frac{x+1}{2x}=\frac{-2}{2x}=\frac{-1}{x}$

Therefore, $h(x)=\frac{-1}{x}$

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