Composition of Functions

A more important way to combine two functions is to form their composition: f ⚬ g.

The definition of composition is given in KEY FACT N6.

Key Fact N6

The composition of two functions f and g is (f ⚬ g)(x) = f(g(x)).

The composition of f and g can be formed only if some of the numbers in the range of g are in the domain of f.

EXAMPLE 12: If $f\left(x\right)=\sqrt{x}$ and g(x) = –x² – 2, then (f ⚬ g)(x) does not exist. For every real number x, g(x) is negative, but no negative number is in the domain of f.

REMEMBER

Do not confuse the product fg with the composition f ⚬ g. (fg)(x) = f(x) · g(x)(f ⚬ g)(x) = f(g(x))

EXAMPLE 13: ­Let f(x) = 2x + 3 and g(x) = x² – 1.

To find (f ⚬ g)(3) you have two choices:

1. You can determine f(g(x)) and then plug in x = 3:
   $\begin{array}{cl}\left(f\circ g\right)\left(x\right)& = f\left(g(x)\right)=2\left(g\left(x\right)\right)+3=2(x^2-1)+3=2x^2-2+3=2x^2+1\\ & \mathrm{Then} \left(\mathrm{f}\circ \mathrm{g}\right)\left(3\right)=2(3^2)+1=2\left(9\right)+1=19\end{array}$

or

1. You can calculate (f ⚬ g)(3) directly:
   
   
   $\left(f\circ g\right)\left(3\right)=f\left(g\left(3\right)\right)=f\left(8\right)=2\left(8\right)+3=19$

EXAMPLE 14: ­Let f and g be the same functions as in Example 13:
f(x) = 2x + 3 and g(x) = x² – 1. Then:

$\begin{array}{rll}\left(g\circ f\right)\left(x\right)& =& g\left(f\left(x\right)\right)=g\left(2x+3\right)={\left(2x+3\right)}^2-1\\ & =& (4x^2+12x+9)-1=4x^2+12x+8\\ & =& 4(x^2+3x+2)=4\left(x+1\right)\left(x+2\right)\end{array}$

$\mathrm{So}, \left(g\mathit{\circ }f\right)\left(3\right)=4\left(3+1\right)\left(3+2\right)=4\left(4\right)\left(5\right)=80$

Of course, (g ⚬ f)(3) could have been calculated directly:

$\left(g\mathit{\circ }f\right)\left(3\right)=g\left(f\left(3\right)\right)\mathit{=}g\left(9\right)\mathit{=}{9}^2-1=80$

Notice that, in general, (f ⚬ g) ≠ (g ⚬ f). In Examples 13 and 14, you saw:

$\left(f\mathit{\circ }g\right)\left(3\right)=19 \mathrm{and} \left(g\mathit{\circ }f\right)\left(3\right)=80$

It is possible, however, to have functions f and g for which (f ⚬ g) = (g ⚬ f).

EXAMPLE 15: ­Let f(x) = 3x – 2 and $g\left(x\right)={\displaystyle \frac{x+2}{3}}$. Then:

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)=f\left(\frac{x+2}{3}\right)=3\left(\frac{x+2}{3}\right)-2=x$

$\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(3x-2\right)=\frac{\left(3x-2\right)+2}{3}=x$