Lesson 53: Performing Compositions of Functions

New Concept

The composite function $f(g(x))$ uses output values from $g(x)$ as input values for $f(x)$. The composition is written $f(g(x))$ or $(f \circ g)(x)$.

What’s next

Next, you’ll practice building and evaluating these composite functions and see why the order of operations is so critical.

Property

If $f(x)$ and $g(x)$ are functions, the composite function $f(g(x))$ uses the output values from $g(x)$ as the input values for $f(x)$. The composition is written as $f(g(x))$ or $(f \circ g)(x)$. It models a multi-step process where the output of one function becomes the input for the next.

Let $f(x) = 4x+3$ and $g(x) = x^2$. Then $(f \circ g)(x) = f(g(x)) = f(x^2) = 4(x^2) + 3 = 4x^2 + 3$.

Let $f(x) = \sqrt{x}$ and $g(x) = x-7$. Then $(f \circ g)(x) = f(g(x)) = f(x-7) = \sqrt{x-7}$.

Property

To evaluate a composite function like $(f \circ g)(a)$, you can use two methods. Method 1: First calculate $g(a)$, then use that result as the input for $f$. Method 2: First find the general composite function $(f \circ g)(x)$, then substitute $x=a$ into the new function.

Let $f(x) = x^2+2$ and $g(x)=2x$. Find $(f \circ g)(3)$.
Method 1: Find $g(3) = 2(3) = 6$. Then find $f(6) = 6^2 + 2 = 38$. So $(f \circ g)(3) = 38$.
Method 2: Find $f(g(x)) = f(2x) = (2x)^2 + 2 = 4x^2 + 2$. Then plug in $x=3$: $4(3)^2 + 2 = 38$.

To solve $(f \circ g)(5)$, you can either plug 5 into $g$, get a result, and plug that result into $f$ (a two-step journey). Or, you can first build the mega-function $f(g(x))$ by mushing them together, and then plug 5 into your powerful new creation. Both paths lead to the same treasure at the end!

Property

For most functions $f(x)$ and $g(x)$, the order of composition matters. This means that performing $f(g(x))$ is not the same as performing $g(f(x))$. In mathematical terms, function composition is not commutative: $f(g(x)) \neq g(f(x))$.

Let $f(x) = x+5$ and $g(x) = 3x$.
$(f \circ g)(x) = f(3x) = 3x + 5$.
$(g \circ f)(x) = g(x+5) = 3(x+5) = 3x + 15$.
Clearly, $3x+5 \neq 3x+15$, so $(f \circ g)(x) \neq (g \circ f)(x)$.

Order is everything! Putting on your socks and then your shoes, $s(h(\text{feet}))$, is a sensible plan that gets you ready for the day. But what about shoes first, then socks, $h(s(\text{feet}))$? That's just a lumpy mess! Function composition is the same way; switching the order of $f(x)$ and $g(x)$ will almost always give you a completely different result.

Property

Composite functions can model real-world situations involving multiple sequential steps, such as applying several discounts to a price. If a sale discount is represented by $s(p)$ and an employee discount by $d(p)$, the final price is found with the composite function $d(s(p))$, where the sale price becomes the input for the employee discount.

A store offers a 15% discount, $s(p)=0.85p$. Employees get an additional 50 dollars off, $e(p)=p-50$. The final price for an employee is $e(s(p)) = e(0.85p) = 0.85p - 50$.

Imagine a double discount on a new video game! First, the store applies a sale function, say 20% off. Then, you use a coupon for another 10 dollars off the sale price. We can combine these two steps into one 'super-function' by composing them, $c(s(p))$, which lets you calculate your amazing final price directly from the original price tag.