EVALUATION OF COMPOSITE FUNCTION

A composite function is a function that is written inside another function. It simply means the outer function depends on the inner function.

Composition of function is done by substituting one function into another function. That is, the composition of function is simply created when one function is substituted into another function.

For example,

$f[g(x)]$, read as ''f of g of x'', is the composite function that is formed when $g(x)$ is substituted for x in $f(x)$.

$f(x)$ is called the outer function while $g(x)$ is called the inner function.

A composite function is usually written using this notation:

$(f\circ g)(x)=f[g(x)]$

The open circle $\circ$ is referred to as the composition operator.

To evaluate a composite function, we first evaluate the value of the inner function and then use the output of the inner function as the input of the outer function. We illustrate these statements with these five examples.

Example 1

Given that $f(x)=x^2$ and $g(x)$=$\frac{1}{x+2}$, Evaluate $f[g(x)]$

Solution:

$f[g(x)]$ means we should plug in $g(x)$ for $x$ in $f(x)$

$f[g(x)]$=$(\frac{1}{x+2})^2$

$f[g(x)]$=$\frac{1}{x+2}\times$$\frac{1}{x+2}$

$f[g(x)]$=$\frac{1}{x(x+2)+2(x+2)}$

when we expand the denominator, we have:

$f[g(x)]=\frac{1}{x^2+4x+4}$

Example 2

If $f(x)=\frac{x+6}{x+2}$ and $g(x)=7-2x^2$, Evaluate $f[g(x)]$

Solution:

Like the first, we plug in the $g(x)$ for $x$ in $f(x)$

$f[g(x)]=\frac{(7-2x^2)+6}{(7-2x^2)+2}$

$f[g(x)]=\frac{-2x^2+6+7}{-2x^2+2+7}$

$f[g(x)]=\frac{-2x^2+13}{-2x^2+9}$

$f[g(x)]=\frac{13-2x^2}{9-2x^2}$

Example 3

Given that $g(x)=x+2$, Evaluate $g[g(x)]$

Solution:

We simply plug in $g(x)$ for $x$ in $g(x)$

$g[g(x)]=(x+2)+2$

$g[g(x)]=x+4$

Example 4

Given $f(x)=2x^2+3x-1$ and $g(x)=3x-4$. Evaluate:

i. $g[f(x)]$

ii) $g^{-1}[f(x)]$

Solution:

i. $g[f(x)]=3(2x^2+3x-1)-4$

$g[f(x)]=6x^2+9x-3-4$

$g[f(x)]=6x^2+9x-7$

ii. To solve $g^{-1}[f(x)]$, we first need to find the inverse of $g(x)$, accordingly

$g(x)=3x-4$.

$y=3x-4$

So, we switch $x$ and $y$

$x=3y-4$

$x+4=3y$

$\frac{x+4}{3}=\frac{3y}{3}$

$y=\frac{x+4}{3}$

Back-substituting $g^{-1}$ for $y$

$g^{-1}(x)=\frac{x+4}{3}$

Having obtained $g^{-1}$, let's now evaluate the composite function $g^{-1}[f(x)]$

$g^{-1}[f(x)]=\frac{x+4}{3}$

$g^{-1}[f(x)]=\frac{(2x^2+3x-1)+4}{3}$

$g^{-1}[f(x)]=\frac{2x^2+3x+3}{3}$

Example 5

Given that $f(x)=4-x^2$ and $g(x)=\frac{5x}{2x-1}$, evaluate $f[g^{-1}(3)]$

Solution

$f[g^{-1}(3)]$ means we should substitute 3 in the inverse of $g(x)$ and then use the result to solve for $f(x)$

However, we are not given the inverse of $g(x)$, so, let quickly solve for the inverse

$g(x)=\frac{5x}{2x-1}$

As usual, we substitute $y$ for $g(x)$

$y=\frac{5x}{2x-1}$

Now, let switch $x$ and $y$

$x=\frac{5y}{2y-1}$

To make y the subject of the formula, we cross multiply

$x(2y-1)=5y$

$2xy-x=5y$

$2xy-5y=x$

$y(2x-5)=x$

$\frac{\require{cancel}\bcancel{2x-5}y}{\require{cancel}\bcancel{2x-5}}=\frac{x}{2x-5}$

$y=\frac{x}{2x-5}$

$g^{-1}(x)=\frac{x}{2x-5}$

Having gotten the inverse of $g(x)$, let now solve for $g^{-1}(3)$

$g^{-1}(x)=\frac{x}{2x-5}$

$g^{-1}(x)=\frac{3}{2(3)-5}$

$g^{-1}(x)=\frac{3}{6-5}$

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

$g^{-1}(x)=3$

Now, let the value of  $g^{-1}(x)$ in $f[g^{-1}(3)]$

$f[3]=4-(3)^2$

$f[3]=4-9$

$f[3]=-5$