### Learning Outcomes

- Decompose a composite function into its component functions

In the Substitution and Integrals Involving Exponential and Logarithmic Functions sections, we will learn all about using substitution as an integration method. Substitution is basically the process used to find the antiderivative of a function that was differentiated using the chain rule. That being said, it is important to be able to look at a composite function and identify the inside function and outside function. Usually, the inside function is what we set our substitution variable equal to.

## Identify Components of Composite Functions

In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions.

### Example: Identifying Components of Composite Functions

Write [latex]f\left(x\right)=\sqrt{5-{x}^{2}}[/latex] as the composition of two functions.

Show Solution
We are looking for two functions, [latex]g[/latex] and [latex]h[/latex], so [latex]f\left(x\right)=g\left(h\left(x\right)\right)[/latex]. To do this, we look for a function inside a function in the formula for [latex]f\left(x\right)[/latex]. As one possibility, we might notice that the expression [latex]5-{x}^{2}[/latex] is the inside of the square root. We could then decompose the function as

[latex]h\left(x\right)=5-{x}^{2}\hspace{2mm}\text{and}\hspace{2mm}g\left(x\right)=\sqrt{x}[/latex]

We can check our answer by recomposing the functions.

[latex]g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}[/latex]

### Example: Identifying Components of Composite Functions

Write [latex]f\left(x\right)=e^{4x-3}[/latex] as the composition of two functions.

Show Solution
We are looking for two functions, [latex]g[/latex] and [latex]h[/latex], so [latex]f\left(x\right)=g\left(h\left(x\right)\right)[/latex]. To do this, we look for a function inside a function in the formula for [latex]f\left(x\right)[/latex]. As one possibility, we might notice that the expression [latex]4x-3[/latex] is within the exponent of the exponential function. We could then decompose the function as

[latex]h\left(x\right)=4x-3\hspace{2mm}\text{and}\hspace{2mm}g\left(x\right)=e^{x}[/latex]

We can check our answer by recomposing the functions.

[latex]g\left(h\left(x\right)\right)=g(4x-3)=e^{4x-3}[/latex]

### Try It

Write [latex]f\left(x\right)=\dfrac{4}{3-\sqrt{4+{x}^{2}}}[/latex] as the composition of two functions.

Show Solution
Possible answer:

[latex]g\left(x\right)=\sqrt{4+{x}^{2}}[/latex]

[latex]h\left(x\right)=\dfrac{4}{3-x}[/latex]

[latex]f=h\circ g[/latex]