407 493 6601

# Algebra 1 - Composite Functions

## Introduction

• Composite Function - Given the functions  the composite function is denoted by $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
• Composition of function is like combining two functions to produce a new function using the concept of evaluating function (direct substitution of x-values).
• The notation $\left(f\circ g\right)\left(x\right)$ is read as "
• The notation $\left(g\circ f\right)\left(x\right)$ is read as ""
• In the composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$, the inner function or independent function is $g\left(x\right)$, while the dependent function or outer function is $f\left(x\right).$
• In the case of $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right),$ the dependent function is $g\left(x\right)$, and the independent function is $f\left(x\right)$.
• The domain of $\left(f\circ g\right)\left(x\right)$ is the set of all real numbers $x$ in the domain of $g$ wherein $g\left(x\right)$ is the domain of $f\left(x\right).$
• The domain of $\left(g\circ f\right)\left(x\right)$ is the set of all real numbers $x$ in the domain of $f$ wherein $f\left(x\right)$ is the domain of $g\left(x\right)$.

## Solved Examples

Example 1. Given that find $\left(f\circ g\right)\left(x\right)$.

Solution:

The composite function $\left(f\circ g\right)\left(x\right)$ is the same as $f\left(g\left(x\right)\right)$. The dependent function is $f$, and the independent function is $g$.

$f\left(g\left(x\right)\right)=f\left(5x+7\right)=3\left(5x+7\right)-4$

$f\left(5x+7\right)=15x+17$

Example 2. If $g\left(x\right)={x}^{2}-7x$, find $\left(g\circ g\right)\left(x\right)$.

Solution:

The dependent function and independent function are the same.

$\left(g\circ g\right)\left(x\right)=g\left(g\left(x\right)\right)$

$g\left({x}^{2}-7x\right)={\left({x}^{2}-7x\right)}^{2}-7\left({x}^{2}-7x\right)$

$g\left({x}^{2}-7x\right)={x}^{4}-14{x}^{3}+49{x}^{2}-7{x}^{2}+49x$

$g\left({x}^{2}-7x\right)={x}^{4}-14{x}^{3}+42{x}^{2}+49x$

Example 3. Given that , find $\left(h\circ g\right)\left(x\right)+\left(g\circ f\right)\left(x\right)$.

Solution:

$\left(h\circ g\right)\left(x\right)=h\left(g\left(x\right)\right)=h\left(5-3x\right)=\left(5-3x\right)-4=1-3x$

$\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left({x}^{2}+9\right)=5-3\left({x}^{2}+9\right)=5-3{x}^{2}-27=-3{x}^{2}-22$

$\left(h\circ g\right)\left(x\right)+\left(g\circ f\right)\left(x\right)=\left(1-3x\right)+\left(-3{x}^{2}-22\right)=-3{x}^{2}-3x-21$

Example 4. If , find $g\left(f\left(-2\right)\right)$.

Solution:

We need to first evaluate .

$f\left(-2\right)=3{\left(-2\right)}^{2}+5=17$

$g\left(17\right)=4\left(17\right)-1=67$

Example 5. Given that , find the ordered pair for the function $\left(f\circ g\right)\left(x\right)$. What is the domain of $\left(f\circ g\right)\left(x\right)$?

Solution:

$f\left(g\left(2\right)\right)=f\left(4\right)=3\to \left(2,3\right)$

$f\left(g\left(3\right)\right)=f\left(3\right)=4\to \left(3,4\right)$

$f\left(g\left(4\right)\right)=f\left(3\right)=4\to \left(4,4\right)$

Thus, the composite function $\left(f\circ g\right)\left(x\right)$ is $\left\{\left(2,2\right),\left(3,4\right),\left(4,4\right)\right\}$, and its domain is $D:\left\{2,3,4\right\}$.

## Solved Examples (Decomposition of Functions)

To decompose a function, we need to determine what functions (independent or dependent) compose a given composite function. Then, we have to break the new function into two functions.

Example 1. If , find $g\left(x\right)$.

Solution:

The unknown here is the inner function $g\left(x\right)$.

Let the composite function be equal to the dependent function .

$f\left(g\left(x\right)\right)=4{x}^{2}+x-2$

$4{x}^{2}+x-2=g\left(x\right)+4$

Thus, the independent function is $g\left(x\right)=4{x}^{2}+x-6$.

Example 2. If , find $f\left(x\right)$.

Solution:

Make an equation for x in terms of $g\left(x\right)$.

$g\left(x\right)=5x-3\to x=\frac{g\left(x\right)+3}{5}$

We substitute $x=\frac{g\left(x\right)+3}{5}$ to $f\left(g\left(x\right)\right)=10{x}^{2}+5x+1$

$f\left(\frac{g\left(x\right)+3}{5}\right)=10{\left(\frac{g\left(x\right)+3}{5}\right)}^{2}+5\left(\frac{g\left(x\right)+3}{5}\right)+1$

Let $x=g\left(x\right)$

$f\left(\frac{x+3}{5}\right)=10{\left(\frac{x+3}{5}\right)}^{2}+5\left(\frac{x+3}{5}\right)+1\to f\left(x\right)=\frac{10{\left(x+3\right)}^{2}}{25}+x+3+1$

$f\left(x\right)=\frac{2{x}^{2}+17x+38}{5}$

## Cheat Sheet

• Evaluating functions is an important skill to master to find the composition of two functions correctly.
• If , then we find the composite function $\left(f\circ g\right)\left(x\right)$ as $f\left(g\left(x\right)\right)=f\left(x+1\right)={\left(x+1\right)}^{3}+2{\left(x+1\right)}^{2}-\left(x+1\right)-1$. Algebraic procedures must be followed to determine the composite function correctly.
• In example like , the domain of $\left(h\circ f\right)\left(x\right)$ is $x\ge -\frac{1}{2}$, and the range of $\left(h\circ f\right)\left(x\right)$ is the range of $h\left(x\right)=\sqrt{x}$.

## Blunder Areas

• In its broad sense, the composition of functions is not commutative. This means that $\left(f\circ g\right)\left(x\right)$ is not always equal to $\left(g\circ f\right)\left(x\right)$. There are only selected cases where commutativity is applicable.