Mathematical-Reasoning · Unit 23 · Video 2 · Interactive Practice

g ∘ f: The Function on the Right Acts First

IKey Formulas

Statement	Meaning
$(g \circ f)(x) = g\big(f(x)\big)$ for all $x \in A$	Definition of the composite $g \circ f : A \to C$ , for $f : A \to B$ and $g : B \to C$
$g \circ f$ : $f$ inner (first), $g$ outer (second)	The function written on the right acts first; the middle set $B$ does not appear in $g \circ f : A \to C$
$(g \circ f)(x) = \sin(3x^2 + 2)$ , $\ (f \circ g)(x) = 3\sin^2 x + 2$	For $f(x) = 3x^2 + 2$ and $g(x) = \sin x$ : in general $g \circ f \ne f \circ g$
$g \circ g$ is defined when $g : B \to B$	A function can be composed with itself when its codomain equals its domain, so every output is again an input

Key Insight: Read $g \circ f$ from right to left: apply $f$ , then feed its output to $g$ . The composite exists because the codomain of $f$ is exactly the domain of $g$ .

IIInner First, Then Outer

With $f(x) = 3x^2 + 2$ and $g(x) = \sin x$ , the two orders of composition give different functions.

x 0

IIIFollowing the Arrows

The composite $g \circ f$ collapses each two-step path $x \to f(x) \to g(f(x))$ into one arrow from $A$ to $C$ .

💡 Challenge: no choice of arrows for $f$ and $g$ makes all four elements of $C$ outputs of $g \circ f$ . Why not?

IVElement by Element, and When a Composite Exists

Each composite is computed one input at a time, and it exists only when the inner function's outputs are legal inputs for the outer one.

Composite Input 0

VQuiz Questions

Problem 1 · Evaluate a Composite

Given: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = 2x + 1$ , and $g : \mathbb{R} \to \mathbb{R}$ , $g(x) = x^2$ — find $(g \circ f)(3)$ .

✅ Correct! The inner function

f

acts first:

f(3) = 7

, then

g(7) = 49

❌ Close, but that is the other order.

19 = f(g(3)) = (f \circ g)(3)

. In

g \circ f

the function on the right,

f

, is applied first.

❌ Not quite.

63 = f(3) \cdot g(3) = 7 \cdot 9

is the product

fg

, not the composite. Composition feeds the output of

f

into

g

❌ Not quite. Compute

f(3)

first, then apply

g

to that output.

Show solution

By the definition of composition, $(g \circ f)(3) = g\big(f(3)\big)$ . First the inner function:

f(3) = 2(3) + 1 = 7

Then the outer function takes that output as its input:

(g \circ f)(3) = g(7) = 7^2 = 49

For comparison, the other order gives $(f \circ g)(3) = f(9) = 19$ , so $g \circ f \ne f \circ g$ here.

Problem 2 · The Formula for a Composite

Given: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^2 - 1$ , and $g : \mathbb{R} \to \mathbb{R}$ , $g(x) = \cos x$ — find a formula for $(f \circ g)(x)$ .

✅ Correct!

g

is the inner function:

f(\cos x) = (\cos x)^2 - 1

❌ Close, but the order is reversed.

\cos(x^2 - 1) = g\big(f(x)\big) = (g \circ f)(x)

. In

f \circ g

g

acts first.

❌ Not quite. That is the product

f(x) \cdot g(x)

. Composition substitutes the whole output

g(x)

for

x

in the formula of

f

❌ Not quite.

\cos(x^2) - 1

squares

x

before taking the cosine, so

g

is not acting first. Squaring must apply to the output

\cos x

Show solution

By definition, $(f \circ g)(x) = f\big(g(x)\big)$ . Replace $g(x)$ by its formula:

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

Now $f$ squares whatever it receives and subtracts $1$ :

(f \circ g)(x) = (\cos x)^2 - 1 = \cos^2 x - 1

The other order is $(g \circ f)(x) = g(x^2 - 1) = \cos(x^2 - 1)$ , a different function: at $x = 0$ it gives $\cos(-1) \approx 0.54$ , while $(f \circ g)(0) = 1 - 1 = 0$ .

Problem 3 · Composites on Finite Sets

Given: $A = \{a, b, c\}$ and $B = \{1, 2, 3\}$ , with $f : A \to B$ defined by $f(a) = 3$ , $f(b) = 1$ , $f(c) = 3$ , and $g : B \to B$ defined by $g(1) = 2$ , $g(2) = 3$ , $g(3) = 1$ — find the arrows of $g \circ f$ and the value $(g \circ g)(3)$ .

What are $(g \circ f)(a)$ , $(g \circ f)(b)$ , $(g \circ f)(c)$ , in that order?

What is $(g \circ g)(3)$ ?

✅ Correct!

g \circ f

sends

a \mapsto 1

b \mapsto 2

c \mapsto 1

, and

(g \circ g)(3) = g(1) = 2

❌ Not quite. These are the values of

f

alone. Each output of

f

must still go through

g

g(3) = 1

and

g(1) = 2

❌ Not quite. These follow the arrows of

g

backwards:

g(2) = 3

means

g

sends

2

3

, so

g

sends

3

1

, not to

2

❌ Check the arrows of $g \circ f$ . For each letter, find

f

of it, then apply

g

to that number.

❌ Not quite.

1 = g(3)

applies

g

only once.

(g \circ g)(3) = g\big(g(3)\big)

needs a second step.

❌ Not quite.

g : B \to B

has codomain equal to its domain,

B

, so

g \circ g

is defined.

❌ Check $(g \circ g)(3)$ . Follow the arrows of

g

twice, starting at

3

Show solution

Apply the definition to each element of $A$ , inner function first:

(g \circ f)(a) = g\big(f(a)\big) = g(3) = 1

(g \circ f)(b) = g\big(f(b)\big) = g(1) = 2

(g \circ f)(c) = g\big(f(c)\big) = g(3) = 1

Both $a$ and $c$ land on $1$ because $f$ already sent both of them to $3$ . Since $g : B \to B$ , the composite $g \circ g$ is defined, and

(g \circ g)(3) = g\big(g(3)\big) = g(1) = 2

Problem 4 · Which Composites Exist?

Given: the disjoint sets $A = \{1, 2, 3\}$ , $B = \{a, b\}$ , $C = \{x, y\}$ and functions $f : A \to B$ , $g : B \to C$ , $h : C \to A$ — decide which composite is not defined, and find the domain and codomain of $f \circ h \circ g$ .

Which composite is NOT defined?

What is $f \circ h \circ g$ ?

✅ Correct!

h \circ f

would feed outputs of

f

(in

B

) to

h

, which accepts only

C

. And

g

acts first, so

f \circ h \circ g

starts in

B

and ends in the codomain of

f

, which is

B

❌ Not that one. That composite is defined: check that the codomain of its right-hand function is the domain of its left-hand function.

❌ Not quite.

A

is the domain of

f

, but

f

is applied last. The domain of a composite is the domain of the right-most, first-acting function.

❌ Check the ends. The composite starts in the domain of the function applied first and ends in the codomain of the function applied last.

Show solution

A composite $q \circ p$ is defined when the codomain of the inner function $p$ is the domain of the outer function $q$ . Check each option:

$f \circ h$ : $h : C \to A$ and $f : A \to B$ , so $f \circ h : C \to B$ is defined.
$h \circ g$ : $g : B \to C$ and $h : C \to A$ , so $h \circ g : B \to A$ is defined.
$g \circ f$ : $f : A \to B$ and $g : B \to C$ , so $g \circ f : A \to C$ is defined.
$h \circ f$ : $f$ outputs elements of $B$ , but $h$ accepts only elements of $C$ , and $B \cap C = \emptyset$ . So $h \circ f$ is not defined.

For $f \circ h \circ g$ , the functions act right to left: $g$ first, then $h$ , then $f$ :

B \xrightarrow{\ g\ } C \xrightarrow{\ h\ } A \xrightarrow{\ f\ } B

Each step lands in the domain of the next, so $f \circ h \circ g : B \to B$ .

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