Example: Finding the Domain of a Composite Function

Find the domain of

[latex]\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\dfrac{5}{x - 1}\text{ and }g\left(x\right)=\dfrac{4}{3x - 2}[/latex]

Answer: The domain of [latex]g\left(x\right)[/latex] consists of all real numbers except [latex]x=\frac{2}{3}[/latex], since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\left(x\right)[/latex] that value of [latex]x[/latex] for which [latex]g\left(x\right)=1[/latex].

[latex]\begin{align}&\dfrac{4}{3x - 2}=1\hspace{5mm}&&\text{Set}\hspace{2mm}g(x)\hspace{2mm}\text{equal to 1} \\[2mm]& 4=3x - 2 &&\text{Multiply by}\hspace{2mm} 3x-2\\[2mm]& 6=3x&&\text{Add 2 to both sides}\\[2mm]& x=2&&\text{Divide by 3} \end{align}[/latex]

So the domain of [latex]f\circ g[/latex] is the set of all real numbers except [latex]\frac{2}{3}[/latex] and [latex]2[/latex]. This means that

[latex]x\ne \frac{2}{3}\hspace{2mm}\text{or}\hspace{2mm}x\ne 2[/latex]

We can write this in interval notation as

[latex]\left(-\infty ,\frac{2}{3}\right)\cup \left(\frac{2}{3},2\right)\cup \left(2,\infty \right)[/latex]

Example: Finding the Domain of a Composite Function Involving Radicals

Find the domain of

[latex]\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\sqrt{x+2}\text{ and }g\left(x\right)=\sqrt{3-x}[/latex]

Answer: Because we cannot take the square root of a negative number, the domain of [latex]g[/latex] is [latex]\left(-\infty ,3\right][/latex]. Now we check the domain of the composite function

[latex]\left(f\circ g\right)\left(x\right)=\sqrt{3-\sqrt{x+2}}[/latex]

For [latex]\left(f\circ g\right)\left(x\right)[/latex], we need [latex]3-\sqrt{x+2}\ge{0}[/latex], again because we cannot take the square root of a negative number. Since the output of a square root is always non-negative, when we add 2, we see that [latex]3-\sqrt{x+2}>0[/latex], so the domain of [latex]\left(f\circ g\right)\left(x\right) = (-\infty,3][/latex].

Analysis of the Solution

This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\circ g[/latex] can contain values that are not in the domain of [latex]f[/latex], though they must be in the domain of [latex]g[/latex]. You cannot rely on an algorithm to find the domain of a composite function. Rather, you will need to first ask yourself "what is the domain of the inner function", and determine whether this set will comply with the domain restrictions of the outer function. In this case, the set [latex](-\infty,3][/latex] ensures a non-negative output for the inner function, which will in turn ensure a positive input for the composite function.

Example: Decomposing a Function

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

Answer: 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]

Analysis of the Solution

For every composition there are infinitely many possible function pairs that will work. In this case, another function pair where [latex]g\left(h\left(x\right)\right)=\sqrt{5-{x}^{2}}[/latex]  is  [latex]h(x)=x^2[/latex] and [latex]g(x)=\sqrt{5-x}[/latex]