Chapter 1 · Relations and Functions

Example 16: $g\circ f \ne f\circ g$ for $\cos x$ and $3x^2$

ExamplesComposition of functions · Section 1.4

The question

Find $gof$ and $fog$ ⁠, if $f : \R \to \R$ and $g : \R \to \R$ are given by $f(x) = \cos x$ and $g(x) = 3x^2$ ⁠. Show that $gof \ne fog$ ⁠.

Exam answer

For $x \in \R$ ⁠,

\begin{aligned} gof(x) &= g(f(x)) = g(\cos x) \\ &= 3\cos^2 x \end{aligned}

\begin{aligned} fog(x) &= f(g(x)) = f(3x^2) \\ &= \cos(3x^2) \end{aligned}

At $x = 0$ ⁠: $gof(0) = 3\cos^2 0 = 3$ and $fog(0) = \cos 0 = 1$ ⁠.

$\therefore\ gof(0) \ne fog(0)$ ⁠, so $gof \ne fog$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The order of composition matters. $gof$ means "do $f$ first, then $g$ ", and $fog$ means "do $g$ first, then $f$ ". Here that is the difference between cosine then square-and-triple, giving $3\cos^2 x$ ⁠, and square-and-triple then cosine, giving $\cos(3x^2)$ ⁠. Two functions are equal only if they agree at every input, so to show they are different we need just one input where they disagree. $x = 0$ does it: $3$ against $1$ ⁠.

Given, to find and show, and the plan

Given: $f(x) = \cos x$ and $g(x) = 3x^2$ ⁠, both from $\R$ to $\R$ ⁠.
To find and show: Formulas for $gof(x)$ and $fog(x)$ ⁠; then that $gof$ and $fog$ are not the same function.
Plan: (1) Check both composites are defined. (2) Substitute to get $gof(x) = g(\cos x)$ ⁠. (3) Substitute to get $fog(x) = f(3x^2)$ ⁠. (4) Recall what "equal functions" means, and find one $x$ where the two values differ.

Basics you need

1Composition

For $f : A\to B$ and $g : B \to C$ ⁠, $gof(x) = g(f(x))$ ⁠. The function nearer to $x$ acts first. Basics: what a function is

2Substituting into a formula

$g(x) = 3x^2$ means "whatever goes in, square it and multiply by 3". So $g(\text{anything}) = 3(\text{anything})^2$ ⁠. The letter $x$ is only a placeholder.

3The notation $\cos^2 x$

$\cos^2 x$ means $(\cos x)^2$ ⁠, the square of the cosine. It is not $\cos(x^2)$ ⁠, the cosine of the square. For example $\cos^2 0 = 1^2 = 1$ ⁠.

4When are two functions equal?

$h = k$ means they have the same domain, the same codomain, and $h(x) = k(x)$ for every $x$ in the domain. So $h \ne k$ as soon as $h(x_0) \ne k(x_0)$ for a single $x_0$ ⁠. Basics: proving and disproving

5A few values of cosine

$\cos 0 = 1$ ⁠, $\cos\frac{\pi}{2} = 0$ ⁠, $\cos \pi = -1$ ⁠, and $-1 \le \cos t \le 1$ for every real $t$ ⁠.

Solution

Both composites are defined

$f$ and $g$ both go from $\R$ to $\R$ ⁠. The codomain of $f$ is $\R$ ⁠, which is the domain of $g$ ⁠, so $gof : \R \to \R$ is defined. In the same way the codomain of $g$ is the domain of $f$ ⁠, so $fog : \R \to \R$ is defined too.

Definition 8 needs the first function's outputs to be legal inputs for the second. Here every real number is a legal input to both $\cos$ and $3x^2$ ⁠, so both orders work. (In Example 15 only one order was possible; here both are, which is what lets us compare them.)

Definition 8: for $f : A \to B$ and $g : B \to C$ ⁠, $gof : A \to C$ ⁠, $gof(x) = g(f(x))$ ⁠. Lesson: composition of functions
Find $gof$ ⁠: cosine first, then $g$
$gof(x) = g(f(x)) = g(\cos x) = 3(\cos x)^2 = 3\cos^2 x.$
We first replace $f(x)$ by what it equals, $\cos x$ ⁠. That leaves $g(\cos x)$ ⁠: the rule $g$ applied to the number $\cos x$ ⁠. The rule says square the input and multiply by 3, so the input $\cos x$ becomes $3(\cos x)^2$ ⁠. The last equality is only the standard shorthand $(\cos x)^2 = \cos^2 x$ ⁠.

$g(\square) = 3\square^2$ whatever is put in the box (Basic 2). Writing the input in brackets, $3(\cos x)^2$ ⁠, avoids confusion with $3\cos(x^2)$ (Basic 3).
Find $fog$ ⁠: $g$ first, then cosine
$fog(x) = f(g(x)) = f(3x^2) = \cos(3x^2).$
Now the inner function is $g$ ⁠, so $g(x) = 3x^2$ is computed first. Then $f$ ⁠, the cosine, is applied to that number. The square sits inside the cosine this time. The two answers contain the same ingredients (a cosine, a square, a factor 3), but in a different order, and that is exactly the point of the example.

$f(\square) = \cos\square$ ⁠. With $\square = 3x^2$ ⁠, the result is $\cos(3x^2)$ ⁠, not $3\cos^2 x$ and not $3\cos(x^2)$ ⁠.
Show $gof \ne fog$ with one value of $x$

Both functions have domain $\R$ and codomain $\R$ ⁠, so they are equal only if $3\cos^2 x = \cos(3x^2)$ for every real $x$ ⁠. Try $x = 0$ ⁠:
$\begin{aligned} gof(0) &= 3\cos^2 0 = 3\cdot 1^2 = 3,\\ fog(0) &= \cos(3\cdot 0^2) = \cos 0 = 1. \end{aligned}$
Since $3 \ne 1$ ⁠, the two functions disagree at $x = 0$ ⁠. Hence $gof \ne fog$ .

Blue: $gof(x) = 3\cos^2 x$ ⁠. Gold: $fog(x) = \cos(3x^2)$ ⁠. At $x = 0$ the blue curve is at 3 and the gold one at 1. The curves do cross at two points in this window (circled, near $x \approx \pm 1.285$ ⁠), so some inputs happen to give equal values, but a single input with different values is enough to make the functions different.

" $gof = fog$ " is a claim about every real number, so one real number where it fails disproves it. That is why the textbook says only "for $x = 0$ ": a single counterexample is a complete proof that the functions are not equal. We chose $x = 0$ because both cosines are easy there ( $\cos 0 = 1$ ⁠). The graph also shows why we must not test at a crossing point: there the two values agree, and that tells us nothing.

Two functions are equal when they agree at every point of a common domain (Basic 4). One counterexample disproves a "for all" statement. Basics: proving and disproving
A second way to see the difference

Since $-1 \le \cos t \le 1$ for every real $t$ ⁠, the function $fog(x) = \cos(3x^2)$ never takes a value larger than 1. But $gof(0) = 3 > 1$ ⁠. So the value 3 is in the range of $gof$ and not in the range of $fog$ ⁠. Functions with different ranges cannot be equal.

This argument does not even need to compare the two functions at the same input. It looks at the set of values each one produces. Their ranges are $[0,3]$ for $gof$ (because $0 \le \cos^2 x \le 1$ ⁠) and $[-1,1]$ for $fog$ ⁠. Different ranges force different functions.

If $h = k$ ⁠, they produce exactly the same outputs, so their ranges are equal. Contrapositive: different ranges mean $h \ne k$ ⁠.

Result

$gof(x) = 3\cos^2 x, \qquad fog(x) = \cos(3x^2), \qquad x \in \R.$ At $x = 0$ ⁠, $gof(0) = 3$ but $fog(0) = 1$ ⁠, so $gof \ne fog$ ⁠. This agrees with the textbook's solution. Composition of functions is not commutative in general.

Check it with numbers

Another input: $x = \pi$ ⁠. Then $gof(\pi) = 3\cos^2\pi = 3(-1)^2 = 3$ ⁠, while $fog(\pi) = \cos(3\pi^2)$ ⁠, and $3\pi^2 \approx 29.61$ ⁠, so $fog(\pi) \approx -0.234$ ⁠. Again different. And at $x = 1$ ⁠: $gof(1) = 3\cos^2 1 \approx 0.876$ ⁠, $fog(1) = \cos 3 \approx -0.990$ ⁠. The verification script checks these values and also confirms that the two curves really do meet near $x \approx 1.285$ ⁠, where both equal about $0.2385$ ⁠.

Common mistakes

Reading $\cos^2 x$ as $\cos(x^2)$ ⁠. $\cos^2 x = (\cos x)^2$ ⁠. At $x = 0$ both happen to be 1, but at $x = \pi$ ⁠, $\cos^2\pi = 1$ while $\cos(\pi^2) \approx -0.903$ ⁠.
Swapping the meaning of $gof$ and $fog$ ⁠. In $gof$ the function $f$ (written on the right, next to $x$ ⁠) acts first.
Writing $fog(x) = 3\cos(x^2)$ ⁠. The factor 3 belongs to $g$ ⁠, which acts first, so it ends up inside the cosine: $\cos(3x^2)$ ⁠.
Trying to "prove" $\ne$ with algebra. No general manipulation is needed. A single value of $x$ where the outputs differ is a complete proof.
Thinking one equal value means equal functions. The curves cross near $x \approx \pm 1.285$ ⁠. Equality at some points does not give equality as functions.

What to take away

To compose, substitute the inner function into the outer one: $gof(x) = g(f(x))$ ⁠.
In general $gof \ne fog$ ⁠: composition is not commutative. Always keep track of which function acts first.
To show two functions are different, find one input where they differ. To show they are equal, you must argue for every input.

Every fact used, and where it comes from

Fact	Used in	Source
Definition of composition	Steps 1–3	Definition 8, Section 1.4 (Lesson: composition)
$\cos^2 x$ means $(\cos x)^2$	Step 2	Standard notation for powers of trigonometric functions
Equal functions agree at every point	Step 4	Meaning of equality of functions (Basics: what a function is)
$\cos 0 = 1$ ⁠; $-1 \le \cos t \le 1$	Steps 4, 5	Values and range of cosine (Class 11)
One counterexample disproves a "for all"	Step 4	Logic of quantifiers (Basics: proving and disproving)
Numerical values, crossing point, graph coordinates	Check, figure	Recomputed in the site's verification script

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Example 16: g∘f≠f∘gg\circ f \ne f\circ gg∘f=f∘g for cos⁡x\cos xcosx and 3x23x^23x2

Basics you need

Solution

Both composites are defined

Find gofgofgof⁠: cosine first, then ggg

Find fogfogfog⁠: ggg first, then cosine

Show gof≠foggof \ne foggof=fog with one value of xxx

A second way to see the difference

Every fact used, and where it comes from

Example 16: $g\circ f \ne f\circ g$ for $\cos x$ and $3x^2$

Find $gof$ ⁠: cosine first, then $g$

Find $fog$ ⁠: $g$ first, then cosine

Show $gof \ne fog$ with one value of $x$