Lesson 3.4: Composition of Functions

New Concept

Function composition links functions by using the output of one as the input for another, creating a new function: $(f \circ g)(x) = f(g(x))$. You'll learn how to build, evaluate, find the domain of, and decompose these functions.

What’s next

Now, let's put this concept into practice. You'll work through interactive examples using formulas, tables, and graphs to master function composition.

Property

For two functions $f(x)$ and $g(x)$ with real number outputs, we define new functions $f + g$, $f - g$, $fg$, and $\frac{f}{g}$ by the relations

Examples

• Given $f(x) = 3x+2$ and $g(x) = x^2$, the sum is $(f+g)(x) = f(x) + g(x) = (3x+2) + x^2 = x^2+3x+2$.
• Given $f(x) = 2x^2$ and $g(x) = 5x-1$, the difference is $(f-g)(x) = f(x) - g(x) = 2x^2 - (5x-1) = 2x^2-5x+1$.
• Given $f(x) = x^2-16$ and $g(x) = x-4$, the quotient is $(\frac{f}{g})(x) = \frac{x^2-16}{x-4} = \frac{(x-4)(x+4)}{x-4} = x+4$, for $x \neq 4$.

Explanation

Think of this as regular arithmetic, but for functions. For any given input $x$, you find the outputs $f(x)$ and $g(x)$ and then add, subtract, multiply, or divide them just like numbers. This creates a new combined function.

Property

When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input $x$ and functions $f$ and $g$, this action defines a composite function, which we write as $f \circ g$ such that

The domain of the composite function $f \circ g$ is all $x$ such that $x$ is in the domain of $g$ and $g(x)$ is in the domain of $f$.

Examples

• If $f(x) = x+5$ and $g(x) = x^2$, then $f(g(x))$ means we plug $g(x)$ into $f$. So, $f(g(x)) = f(x^2) = x^2+5$.
• If $f(x) = x+5$ and $g(x) = x^2$, then $g(f(x))$ means we plug $f(x)$ into $g$. So, $g(f(x)) = g(x+5) = (x+5)^2 = x^2+10x+25$.
• If $r(t)$ is the radius of a balloon after $t$ seconds and $V(r)$ is the volume for a given radius $r$, then $V(r(t))$ represents the volume of the balloon as a function of time.

Explanation

Function composition is like a chain reaction. The input $x$ goes into function $g$, and its output, $g(x)$, becomes the input for function $f$. The order matters, as $f(g(x))$ is usually different from $g(f(x))$.

Property

To evaluate a composite function, we work from the inside out.

1. Evaluate the inside function using the input value or variable provided.
2. Use the resulting output as the input to the outside function.

Examples

• Given $f(x) = 2x+1$ and $h(x) = x^2$, to evaluate $f(h(3))$, first find $h(3) = 3^2 = 9$. Then use this output as the input for $f$: $f(9) = 2(9)+1 = 19$.
• Given $f(t) = t^2-5$ and $h(x) = 2x$, to evaluate $h(f(2))$, first find $f(2) = 2^2-5 = -1$. Then use this output as the input for $h$: $h(-1) = 2(-1) = -2$.
• From a table where $g(4)=7$ and $f(7)=1$, evaluating $f(g(4))$ means finding $g(4)$ first, which is 7. Then we find $f(7)$, which is 1. So, $f(g(4))=1$.

Explanation

Always start with the innermost function. Calculate its value for the given input first. Then, take that result and use it as the new input for the outer function to get your final answer. It is a two-step process.

Property

The domain of a composite function $f(g(x))$ is the set of those inputs $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$.
To find the domain of a composition $f(g(x))$:

1. Find the domain of $g$.
2. Find the domain of $f$.
3. Find those inputs $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$. Exclude inputs $x$ from the domain of $g$ for which $g(x)$ is not in the domain of $f$.

Examples

• For $f(x) = \frac{1}{x-2}$ and $g(x) = \frac{3}{x-4}$, the domain of $g$ is $x \neq 4$. We also need $g(x) \neq 2$. Solving $\frac{3}{x-4}=2$ gives $x=5.5$. So the domain is $(-\infty, 4) \cup (4, 5.5) \cup (5.5, \infty)$.
• For $f(x) = \sqrt{x-3}$ and $g(x) = 2x+5$, the domain of $g$ is all real numbers. We need the output $g(x)$ to be in the domain of $f$, so $g(x) \ge 3$. This means $2x+5 \ge 3$, which simplifies to $x \ge -1$. The domain is $[-1, \infty)$.
• For $f(x) = \frac{1}{x}$ and $g(x) = x^2-25$, the domain of $g$ is all real numbers. We need $g(x) \neq 0$, so $x^2-25 \neq 0$. This means $x \neq 5$ and $x \neq -5$. The domain is $(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$.

Explanation

To find the domain of a composite function, you must satisfy two conditions. First, the input $x$ must be allowed in the inner function $g$. Second, the output of the inner function, $g(x)$, must be an allowed input for the outer function $f$.

Property

To decompose a function, we write it as a composition of two simpler functions. When looking for a decomposition of $f(x)$ into $g(h(x))$, we search for a function $h(x)$ inside the formula for $f(x)$. Then the outer function $g(x)$ is what remains. There may be more than one way to decompose a composite function.

Examples

• To decompose $f(x) = (2x-7)^4$, we can choose the inner part as $h(x) = 2x-7$. The outer operation is raising to the fourth power, so $g(x) = x^4$.
• To decompose $f(x) = \sqrt{x^2+9}$, we can choose the expression inside the square root as the inner function: $h(x) = x^2+9$. The outer function is then $g(x) = \sqrt{x}$.
• To decompose $f(x) = \frac{5}{x+3}$, we can set the denominator as the inner function, $h(x) = x+3$. The outer function then becomes $g(x) = \frac{5}{x}$.

Explanation

Decomposition is like working backward. You look at a complex function and try to spot an 'inner' piece. This inner piece becomes your inside function, $h(x)$, and the operation being performed on it becomes your outer function, $g(x)$.