1.1.3 Composition and inverse functions

Composition: one function after another

Take $f(x) = 2x + 1$ and $g(x) = x^2$ . Feed 3 into $g$ to get 9, then feed 9 into $f$ to get 19. That two-step machine is the composition of $f$ with $g$ .

Rule: Composition

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

Read $f \circ g$ as " $f$ after $g$ ". Work from the inside out: apply $g$ first, then apply $f$ to the result.

The same two machines in both orders. Squaring then doubling gives 19; doubling then squaring gives 49.

The order matters. $f(g(3)) = f(9) = 19$ , but $g(f(3)) = g(7) = 49$ . A table at a few inputs shows the two compositions are different functions:

$x$	0	1	2	3
$f(g(x))$	1	3	9	19
$g(f(x))$	1	9	25	49

For a formula, substitute the whole inner function, brackets and all. $f(g(x)) = 2x^2 + 1$ , while $g(f(x)) = (2x + 1)^2$ , which multiplies out to $4x^2 + 4x + 1$ . The two formulas agree only at $x = 0$ and $x = -2$ .

Sometimes the order does not matter. Adding 2 and adding 5 give $x + 7$ in either order. That is the exception, so always check.

Try it compose(outer, inner) returns a new function that applies inner and then outer . Functions are values in Python: you can pass them in and hand them back, as the exercise in lesson 1.1.4 does. The two lists are the rows of the table.

def f(x): return 2 * x + 1 def g(x): return x**2 def compose(outer, inner): return lambda x: outer(inner(x)) f_after_g = compose(f, g) g_after_f = compose(g, f) print([f_after_g(x) for x in [0, 1, 2, 3]]) print([g_after_f(x) for x in [0, 1, 2, 3]])

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Units chain together

A car at 80 km per hour has gone $g(t) = 80t$ km after $t$ hours. It burns $f(d) = 0.06d$ litres of fuel over $d$ km. Then $f(g(t)) = 0.06 \times 80t = 4.8t$ litres after $t$ hours.

The units run hours, then km, then litres. After 5 hours the car has gone 400 km and used 24 litres. The other order, $g(f(t))$ , would feed litres into a function that expects hours: it means nothing.

Which inputs a composition accepts

The inner output must be an allowed input of the outer function. Take $f(x) = \sqrt{x}$ and $g(x) = x - 4$ . $g$ accepts every number, but $f(g(x)) = \sqrt{x - 4}$ needs $x - 4 \ge 0$ , so its domain is $x \ge 4$ .

The other order, $g(f(x)) = \sqrt{x} - 4$ , needs only $x \ge 0$ .

This also explains the sideways shifts of lesson 1.1.2. $\sqrt{x - 4}$ is $\sqrt{x}$ applied after the inner function $x - 4$ : a sideways shift is a composition.

x = 0, 4, 8, 13

import numpy as np x = np.array([0.0, 4.0, 8.0, 13.0]) with np.errstate(all="ignore"): print(np.sqrt(x - 4)) print(np.round(np.sqrt(x) - 4, 4))

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Inner and outer

Reading a formula as a composition means finding its last step. Ask what you would compute last if you evaluated it at a number.

$(3x + 1)^5$ : first $3x + 1$ , then the 5th power. The inner function is $3x + 1$ and the outer is $u^5$ , with $u$ standing for whatever the inner gives.
$\sqrt{x^2 + 1}$ : inner $x^2 + 1$ , outer $\sqrt{u}$ .
$2x^2 + 1$ : inner $x^2$ , outer $2u + 1$ . That is $f(g(x))$ from above.

The split is a way of reading, not a new rule. Lesson 2.3.3 uses it to find the slope of a composition, the chain rule.

Inverse functions: running backwards

$f(x) = 2x + 3$ doubles its input and adds 3. To undo it, subtract 3 and then halve: undo the steps in reverse order, as in lesson 0.3.2. That undoing function is the inverse of $f$ , written $f^{-1}$ : here $f^{-1}(x) = \frac{x - 3}{2}$ .

$x$	0	1	2.5	10
$f(x) = 2x + 3$	3	5	8	23
$f^{-1}(f(x))$	0	1	2.5	10

Rule: Inverse function

\begin{gathered} f^{-1}\big(f(x)\big) = x \\ f\big(f^{-1}(y)\big) = y \end{gathered}

The inverse sends each output back to the input it came from. The $-1$ is a label, not a power: $f^{-1}(x)$ is not $\frac{1}{f(x)}$ .

To find a formula, write $y = f(x)$ and make $x$ the subject. For $y = 2x + 3$ : subtract 3, then divide by 2, giving $x = \frac{y - 3}{2}$ . Then, by habit, rename the input back to $x$ .

The reciprocal is a different thing. Here $f^{-1}(5) = 1$ , because $f(1) = 5$ . But $\frac{1}{f(5)} = \frac{1}{13}$ .

Units swap too

The inverse swaps inputs and outputs, units included. The fuel rule $f(d) = 0.06d$ takes km to litres. Its inverse, $f^{-1}(L) = \frac{L}{0.06}$ , takes litres back to km: 30 litres cover 500 km.

Domain and range swap

Lesson 1.1.1 solved $\sqrt{x - 3} = y$ and got $x = y^2 + 3$ . That is the inverse: if $g(x) = \sqrt{x - 3}$ , then $g^{-1}(x) = x^2 + 3$ . For example, $g(7) = 2$ and $g^{-1}(2) = 7$ .

The inverse accepts only what $g$ can produce. $g$ has domain $[3, \infty)$ and range $[0, \infty)$ , so $g^{-1}$ has domain $[0, \infty)$ and range $[3, \infty)$ . The formula $x^2 + 3$ would accept $-2$ , but $g^{-1}$ does not.

Rule: Domain and range of an inverse

\begin{gathered} \text{domain of } f^{-1} = \text{range of } f \\ \text{range of } f^{-1} = \text{domain of } f \end{gathered}

Running backwards turns outputs into inputs, so the two sets trade places.

f^{-1}(f(x))

import sympy as sp def f(x): return 2 * x + 3 def f_inv(y): return (y - 3) / 2 for x in [0, 1, 2.5, 10]: print(x, f(x), f_inv(f(x))) x, y = sp.symbols("x y", real=True) print(sp.solve(sp.Eq(y, 2*x + 3), x)) print(sp.solve(sp.Eq(y, sp.sqrt(x - 3)), x))

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When an inverse exists

An inverse can only exist if no two inputs share an output. $x^2$ sends both $-2$ and 2 to 4, so an inverse would not know which one to send 4 back to.

Rule: Horizontal line test

\begin{gathered} \text{each horizontal line meets} \\ \text{the graph at most once} \end{gathered}

A function has an inverse exactly when no horizontal line crosses its graph twice. Such a function is called one-to-one.

Every straight line that is not flat passes, and so does $x^3$ : $x^3 = 8$ only at 2, and $x^3 = -8$ only at $-2$ . A flat line such as $y = 5$ fails as badly as possible, since every input gives the same output.

A function that fails can often be fixed by shrinking its domain. Keep only $x \ge 0$ and $x^2$ passes, with inverse $\sqrt{x}$ . Keep only $x \le 0$ instead and the inverse is $-\sqrt{x}$ , which sends 9 back to $-3$ .

The horizontal line $y = 4$ meets $x^2$ twice, so $x^2$ has no inverse. Kept to $x \ge 0$ , it does: $\sqrt{x}$ is its mirror image in $y = x$ , and $(2, 4)$ becomes $(4, 2)$ .

The mirror in $y = x$

The graph of $f^{-1}$ is the graph of $f$ mirrored in the line $y = x$ . Swapping inputs and outputs swaps the two coordinates of every point: $(2, 4)$ on $x^2$ becomes $(4, 2)$ on $\sqrt{x}$ . Lesson 1.4.3 meets the same mirror between $e^x$ and $\ln x$ , the logarithm defined as an inverse.

\sqrt{x - 3}

import math def g(x): return math.sqrt(x - 3) def g_inv(x): return x**2 + 3 print(g(7), g_inv(g(7))) print(g_inv(5), g(g_inv(5))) print(2**2, math.sqrt(4))

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Case study 1.3.5 uses an inverse from finance: a bond's yield is read back from its price.

Common mistakes

Composing in the wrong order. $f(g(x))$ applies $g$ first.
Dropping brackets when substituting. $g(f(x)) = (2x + 1)^2$ , not $2x + 1^2$ .
Forgetting the domain of a composition. $\sqrt{x - 4}$ needs $x \ge 4$ , even though $x - 4$ accepts every number.
Reading $f^{-1}$ as a reciprocal. $f^{-1}(x)$ undoes $f$ ; $\frac{1}{f(x)}$ is something else.
Inverting a function that is not one-to-one. Shrink the domain first, and say which part you kept.
Keeping the formula's domain for the inverse. $g^{-1}(x) = x^2 + 3$ accepts only the range of $g$ , $x \ge 0$ .

Composition and inverses

0 of 12 correct

1.
$f(x) = 2x + 1$ and $g(x) = x^2$ . Find $f(g(3))$
2.
With the same $f$ and $g$ , find $g(f(3))$
3.
With the same $f$ and $g$ , write $g(f(x))$ as a formula
4.
A car goes $g(t) = 80t$ km in $t$ hours and burns $f(d) = 0.06d$ litres over $d$ km. How many litres is $f(g(5))$ ?
5.
$f(x) = \sqrt{x}$ and $g(x) = x - 4$ . What is the smallest input of $f(g(x))$ ?
6.
In $(3x + 1)^5$ , the outer function is $u^5$ . What is the inner function?
7.
Find the inverse of $f(x) = 3x - 6$ . Write it in terms of $x$
8.
$f(x) = 2x + 3$ . Find $f^{-1}(5)$
9.
The fuel rule $f(d) = 0.06d$ gives litres for $d$ km. How many km does $f^{-1}(30)$ give?
10.
$f(x) = x^3$ . Find $f^{-1}(-8)$
11.
$g(x) = \sqrt{x - 3}$ has inverse $g^{-1}(x) = x^2 + 3$ for $x \ge 0$ . Find $g^{-1}(5)$
12.
$x^2$ is kept to $x \le 0$ . Its inverse sends 9 to which number?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

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

Composition and inverse functions

Composition: one function after another

Rule: Composition

Units chain together

Which inputs a composition accepts

Inner and outer

Inverse functions: running backwards

Rule: Inverse function

Units swap too

Domain and range swap

Rule: Domain and range of an inverse

When an inverse exists

Rule: Horizontal line test

The mirror in $y = x$

Common mistakes

Composition and inverses

Summary card

Free further reading

Composition: one function after another

Rule: Composition

Units chain together

Which inputs a composition accepts

Inner and outer

Inverse functions: running backwards

Rule: Inverse function

Units swap too

Domain and range swap

Rule: Domain and range of an inverse

When an inverse exists

Rule: Horizontal line test

The mirror in y=xy = xy=x

Common mistakes

Summary card

Free further reading

The mirror in $y = x$