Combining Functions

Main idea

You can add, multiply and divide functions just like numbers. You can also chain them: in $f(g(x))$ , whatever comes out of $g$ goes straight into $f$ , and the order you chain them in usually matters.

(f + g)(x) = f(x) + g(x), \qquad (fg)(x) = f(x)\,g(x), \qquad \left(\frac fg\right)(x) = \frac{f(x)}{g(x)}

Composition

(f {\htmlData{sym=0}{\circ}} g)(x) = f({\htmlData{sym=1}{g(x)}})

What the symbols mean

$\circ$
Composition
Apply g first, then f.
$g(x)$
Inner function
Its output is fed into f.

Explanation

Combining functions is simply building a new function out of two old ones, either by doing arithmetic with their outputs or by running one after the other. The first row of formulas just says "do $f$ , do $g$ , then add (or multiply, or divide) the answers." Composition is the interesting one: $(f \circ g)(x) = f(g(x))$ has $g(x)$ (the inner function, which goes first) and $f$ (the outer one, which takes $g$ 's answer as its input). The little circle $\circ$ is read "composed with," and it helps to think of $f \circ g$ as " $f$ after $g$ ." It's like an assembly line, and that's all there is to it.

Order matters. Think about the machines from last lesson. Feeding $x$ through $g$ and then $f$ is a different process from $f$ first and then $g$ , so $f(g(x))$ and $g(f(x))$ are usually different functions. Putting on socks and then shoes isn't the same as shoes and then socks. Always work from the inside out.

Domains. $f + g$ , $fg$ and $\frac fg$ need $x$ to be in both domains, since you have to be able to compute both outputs, and $\frac fg$ also needs $g(x) \ne 0$ . For $f(g(x))$ there are two hurdles: $x$ has to be a legal input for $g$ , and then $g(x)$ has to be a legal input for $f$ .

Decomposition runs the other way. $h(x) = (3x - 1)^4$ is $f(g(x))$ with inner $g(x) = 3x - 1$ and outer $f(x) = x^4$ : first you compute $3x - 1$ , then you raise it to the 4th power. Spotting the inner function like this is the first step of the chain rule in calculus.

See it

A store offers 10 off with a coupon, f(x) = x − 10, and 20% off in a sale, g(x) = 0.8x. Using both is a composition, and the order matters. Taking the 20% first gives f(g(x)) = 0.8x − 10; using the coupon first gives g(f(x)) = 0.8x − 8. As you can see, coupon-first always costs exactly 2 more, because the 20% discount then shrinks the coupon too.

With f(x) = 2x + 1 and g(x) = x², the two orders give different curves: f(g(x)) = 2x² + 1 has its low point at (0, 1), while g(f(x)) = (2x + 1)², drawn dashed, bottoms out at (−1/2, 0).

Worked examples

Example 1

For $f(x) = 2x + 1$ and $g(x) = x^2$ , find $f(g(x))$ and $g(f(x))$ .

1
For $f(g(x))$ , $g$ goes first and gives $x^2$ ; then $f$ doubles it and adds 1: $f(g(x)) = 2(x^2) + 1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

For $f(x) = 2x + 1$ , $g(x) = x^2$ , find $(f \circ g)(x)$ .

Problem 2+5 XP

For the same functions, find $(g \circ f)(x)$ .

Problem 3+5 XP

For the same functions, find $(f \circ g)(3)$ .

Problem 4+5 XP

For $f(x) = \sqrt x$ and $g(x) = x - 5$ , what is the domain of $f(g(x))$ ?

Problem 5+5 XP

$h(x) = (3x - 1)^4$ is $f(g(x))$ with $g(x) = 3x - 1$ . What is $f(x)$ ?

Problem 6+5 XP

A balloon's radius is $r(t) = 2t + 1$ cm after $t$ seconds, and a sphere's surface area is $S(r) = 4\pi r^2$ . Find $S(r(2))$ , the surface area after 2 seconds.

Problem 7+5 XP

For $f(x) = \dfrac1x$ and $g(x) = x^2 - 4$ , which inputs must be excluded from the domain of $f(g(x))$ ?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

If $f(x) = x^2 + 1$ and $f(g(x)) = 4x^2 - 12x + 10$ , find $g(x)$ , choosing the one with a positive leading coefficient.

Where it's used

Pipelines are compositions: a currency conversion after a sales tax, a unit change after a measurement, a neural network's layers each feeding the next.

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