Lesson 5 of 7Rates of Change & Combining Functions

Unit 6 · Lesson 5 of 7

Composition of Functions & Modelling

Feed the output of one function straight into another to get a composite function, then use compositions to model real situations — including ones that can only be solved graphically or numerically.

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Key things to remember

1 $(f \circ g) (x) = f (g (x))$ — the inner function acts first; read the notation from the inside out.
2Composition is not commutative: $f (g (x)) \neq = g (f (x))$ in general.
3Domain of $f \circ g$ : all $x$ in the domain of $g$ whose output $g (x)$ also lies in the domain of $f$ .
4To decompose, find the operation performed last — that becomes the outer function $f$ .
5If $g$ is the inverse of $f$ , then $f (g (x)) = x$ and $g (f (x)) = x$ on the appropriate domains.
6When no algebraic method exists, solve by graphing both sides or by narrowing a sign change numerically, and state the precision of the answer.

A pebble dropped into a pond makes a circular ripple. The radius of the ripple grows with time, and the area of the circle depends on the radius. So the area depends on time — but only indirectly, through the radius. Two separate rules are chained together: time goes in, a radius comes out, that radius goes in, an area comes out. Chaining functions like this is called composition, and it is how almost every real model is actually assembled.

Concentric circular ripples spreading outward across a still, grey water surface. — PhotoRipples spreading in circles from a disturbance in still water. The radius grows with time and the area depends on the radius, the same chain of two functions this lesson opens with.Roger McLassus, CC BY-SA 3.0, via Wikimedia Commons

One machine feeding another

Line up two function machines so the output chute of $g$ empties into the input hopper of $f$ . Put $x$ in, get $g (x)$ out, and feed that straight into $f$ to get $f (g (x))$ . We write this as $f \circ g$ , read " $f$ composed with $g$ " or " $f$ of $g$ of $x$ ". The function written on the inside acts first — reading the notation from the inside out is reading it in the order things actually happen. And because order is built into the notation, swapping the two machines usually changes the result: socks then shoes is a sensible outcome, shoes then socks is not.

(f \circ g) (x) = f (g (x)) (g \circ f) (x) = g (f (x))

Fig 1Two machines in series: the output of the inner function becomes the input of the outer one.

Order matters

AInner function g acts first

f(g(x)) = (2x − 3)² + 1
= 4x² − 12x + 9 + 1
= 4x² − 12x + 10
At x = 2: f(g(2)) = f(1) = 2

BInner function f acts first

g(f(x)) = 2(x² + 1) − 3
= 2x² + 2 − 3
= 2x² − 1
At x = 2: g(f(2)) = g(5) = 7

VideoA full Advanced Functions lesson on composite functions, feeding one function's output into another.Watch on YouTube

The domain of a composition

For $f (g (x))$ to make sense, two things must both be true. First, $x$ has to be a legal input for $g$ — otherwise nothing comes out of the first machine at all. Second, the value $g (x)$ that comes out has to be a legal input for $f$ . Miss either condition and the chain breaks. So you check the inner function's domain, then check which of those survivors produce outputs $f$ will accept.

Check the inner function first

Simplifying before finding the domain hides restrictions. Take

f (x) = (x)^{2}

: the simplified rule looks like

x

, valid everywhere, but the inner square root refuses negatives, so the real domain is

x \geq 0

. The domain always comes from the chain as built, never from the tidy version.

Decomposing a complicated function

The reverse skill is just as useful: given a messy function, spot the simpler ones it is built from. Look for the operation applied last — that outer action is $f$ , and whatever it was applied to is $g$ . There is usually more than one valid answer, so aim for pieces that are as simple as possible.

Composite h(x)	Inner g(x)	Outer f(x)	Last operation performed
$3 x^{2} + 5$	$3 x^{2} + 5$	$x$	take a square root
$(2 x + 7)^{5}$	$2 x + 7$	$x^{5}$	raise to the fifth power
$lo g (x^{2} - 1)$	$x^{2} - 1$	$lo g x$	take a logarithm
$sin^{2} x$	$sin x$	$x^{2}$	square the result
$\frac{1}{x + 4}$	$x + 4$	$\frac{1}{x}$	take a reciprocal

Decomposition: identify the last operation to find the outer function.

VideoWorks backwards from a complicated function to an inner and an outer function whose composition rebuilds it.Watch on YouTube

When algebra runs out

Combining and composing functions produces equations that no algebraic technique can solve. You can solve $2^{x} = 16$ with logarithms and $x^{2} = 16$ with roots, but $x^{2} = 2^{x}$ mixes a polynomial with an exponential, and there is simply no rearrangement that isolates $x$ . This is normal, not a sign you have missed a method. The response is to switch tools: graph both sides and read off the intersections, or evaluate the difference at trial values and narrow down where it changes sign. A modelling answer found this way is a real answer — just state the accuracy you used, for example "correct to two decimal places".

Fig 2Choosing a solution method: exact algebra when a technique exists, otherwise graph or narrow down numerically.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Let

f (x) = x^{2} + 1

and

g (x) = 2 x - 3

. Find

f (g (x))

g (f (x))

f (g (2))

and

g (f (2))

Example 2

Let

f (x) = \frac{1}{x - 2}

and

g (x) = x

. Find

f (g (x))

and state its domain.

Example 3

A store takes 20 dollars off a jacket, then charges 13% HST on the reduced price. Write this as a composition, find the final price of a 250-dollar jacket, and compare it with charging the tax first.

Example 4

Solve

x^{2} = 2^{x}

for all real

x

, to two decimal places where necessary.

Where marks get lost

Composing in the wrong order — computing $g (f (x))$ when the question asked for $f (g (x))$ , or vice versa.
Confusing $(f g) (x)$ , which means $f (x) \cdot g (x)$ , with $(f \circ g) (x)$ , which means $f (g (x))$ .
Dropping brackets when substituting, so $f (g (x)) = (2 x - 3)^{2}$ is written as $2 x - 3^{2}$ .
Finding the domain from the simplified composite instead of from the chain, which erases restrictions the inner function imposed.
Grinding away at algebra on an equation like $x^{3} = 2^{x}$ that has no algebraic solution, instead of switching to a graphical or numerical method.

Flashcards 8

(f \circ g) (x)

Which function acts first?

f (g (x)) = g (f (x))

Domain of a composition

+ 4 more

Practice

Let

f (x) = x^{2} + 1

and

g (x) = 2 x - 3

. Find

f (g (2))

With the same

f (x) = x^{2} + 1

and

g (x) = 2 x - 3

, find

g (f (2))

For

f (x) = x^{2} + 1

and

g (x) = 2 x - 3

, which is the simplified form of

f (g (x))

Match each composite function to the outer function f that would produce it as f(g(x)).

A. $f (x) = \frac{1}{x}$
B. $f (x) = lo g x$
C. $f (x) = x$
D. $f (x) = x^{5}$

h (x) = 4 x + 9

h (x) = (4 x + 9)^{5}

h (x) = \frac{1}{4 x + 9}

h (x) = lo g (4 x + 9)

Let

f (x) = x

and

g (x) = x - 7

. What is the smallest value in the domain of

f (g (x))

Let

f (x) = \frac{1}{x - 3}

and

g (x) = x

. What is the domain of

f (g (x))

A store takes 20 dollars off a 250-dollar jacket, then charges 13% HST on the reduced price. What does the customer pay?

Using the same jacket, what would the customer pay if the 13% HST were charged first and the 20 dollars taken off afterward?

Order the steps for evaluating f(g(a)) for a specific number a:

1Simplify to get the single number g(a)
2Simplify to get the final value f(g(a))
3Substitute that number into the outer function f
4Substitute a into the inner function g

Which method should you use to solve

x^{3} = 2^{x}

For general functions

f

and

g

, does

f (g (x))

always equal

g (f (x))

? Answer yes or no.

A circular oil slick has radius

r (t) = 0.5 t

km after

t

hours, and a circle of radius

r

has area

A (r) = π r^{2}

. Find

A (r (6))

, the area after 6 hours, in square kilometres, rounded to two decimal places.

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