Composition of functions — f(g(x)), g(f(x)): Worked examples

Before anything else, one reminder we will lean on constantly:

Read the figure: the input $x$ enters at the right, flows leftward through machine $g$ (deep teal), whose output $g (x)$ is handed off to machine $f$ (burnt orange), which finally emits $f (g (x))$ at the left. The plum circle in the middle is the composition symbol $\circ$ — the arrow curving right-to-left is your reminder that the rightmost function runs first.

The scenario matrix

Every composition problem is one (or a blend) of these cells. Each example below is tagged with the cell it hits.

Cell	What makes it tricky	Example
A — Plain algebra, both orders	Show $f \circ g \neq = g \circ f$	Ex 1
B — Domain trap (square root / log)	Inner output must fit outer's mouth	Ex 2
C — Zero & degenerate input	Denominator $\to 0$ , undefined points	Ex 3
D — Self-composition $f \circ f$	Same machine twice; watch fixed points	Ex 4
E — Inverse pair (composition = identity)	$f \circ g = x$ signals undo	Ex 5
F — Decompose a given $h$	Split into inner + outer	Ex 6
G — Real-world word problem (units!)	Chain two rate/conversion functions	Ex 7
H — Exam twist: solve $f (g (x)) = k$	Compose, then solve an equation	Ex 8
I — Sign/piecewise & limiting behaviour	Absolute value, quadrants, $x \to \infty$	Ex 9

Prerequisites we quietly use: Domain and range of function, Inverse functions, and for one aside Exponential and logarithmic functions. If chain-rule calculus tempts you, that lives at Derivatives - chain rule — different topic, same nesting idea.

Cell A — Plain algebra, both orders

Worked example Example 1 — worked

Read the figure: the top track (teal $g$ first, then orange $f$ ) is $f \circ g$ ; the bottom track (orange $f$ first, then teal $g$ ) is $g \circ f$ . The input $3$ splits down both tracks and exits at two different numbers — the visual proof that order matters.

Write $(f \circ g) (x) = f (g (x))$ . Inner is $g$ , so substitute $g (x) = x^{2}$ into $f$ : $f (x^{2}) = 2 (x^{2}) - 1 = 2 x^{2} - 1.$ Why this step? The rightmost machine $g$ runs first; its output $x^{2}$ becomes $f$ 's input.
Write $(g \circ f) (x) = g (f (x))$ . Now $f$ runs first: $g (2 x - 1) = (2 x - 1)^{2} = 4 x^{2} - 4 x + 1.$ Why this step? We replace every $x$ in $g$ 's rule ( $x^{2}$ ) with the whole expression $2 x - 1$ .
Evaluate at $x = 3$ . $(f \circ g) (3) = 2 (9) - 1 = 17$ ; $(g \circ f) (3) = 4 (9) - 12 + 1 = 25$ . Why this step? Plugging a number confirms the two orders truly differ, not just look different — and matches the two exit numbers in the figure.

Verify: Trace by hand. $(f \circ g) (3)$ : $g (3) = 9$ , then $f (9) = 17$ ✓. $(g \circ f) (3)$ : $f (3) = 5$ , then $g (5) = 25$ ✓. Different values ⇒ order matters.

Recall Reveal the core rule

In $f (g (x))$ which machine runs first? ::: the inner one, $g$ — the rightmost function always processes the input first.

Cell B — Domain trap

Worked example Example 2 (Cell B)

Given $f (x) = x$ and $g (x) = 5 - x$ . Find the formula and the domain of $(f \circ g) (x)$ .

Forecast: The naive answer is " $x \geq 0$ because square root." Is that right here?

Build the composite: $(f \circ g) (x) = f (5 - x) = 5 - x$ . Why this step? We need the explicit formula before we can hunt for illegal inputs.
List the two gatekeepers. (a) $x$ must be legal for $g$ — always true, subtraction eats any real. (b) The output $g (x)$ must be legal for $f$ — square root needs $\geq 0$ . Why this step? A composite has two doors: the input door of $g$ and the "handoff" door into $f$ .
Impose the handoff condition: $g (x) \geq 0 \Rightarrow 5 - x \geq 0 \Rightarrow x \leq 5$ . Why this step? If $g$ ever outputs a negative, $\cdot$ is undefined there.

$(f \circ g) (x) = 5 - x, domain (- \infty, 5]$

Verify: Test $x = 5$ : $0 = 0$ ✓ (boundary allowed). Test $x = 6$ : $- 1$ undefined ✓ (excluded). The " $x \geq 0$ " guess was wrong — the minus sign flips it.

Read the figure: the number line below shows the two gatekeeper conditions stacked. The teal band is where $g$ accepts input (all $x$ ); the orange band is where the handoff $g (x) \geq 0$ holds ( $x \leq 5$ ). The composite lives only where both bands overlap — the shaded region ending at the plum boundary dot $x = 5$ .

Cell C — Zero & degenerate input

Worked example Example 3 (Cell C)

Given $f (x) = \frac{1}{x}$ and $g (x) = x - 2$ . Find $(f \circ g) (x)$ and the value(s) of $x$ where it blows up.

Forecast: A reciprocal explodes when its input is $0$ . Which $x$ makes $g (x) = 0$ ?

Compose: $(f \circ g) (x) = f (x - 2) = \frac{1}{x - 2}$ . Why this step? Substituting $g$ into $f$ shows the denominator is $x - 2$ , not $x$ .
Find the degenerate input: set denominator $= 0$ : $x - 2 = 0 \Rightarrow x = 2$ . Why this step? Division by zero is undefined; that single point is a hole in the domain.
Describe limiting behaviour near $x = 2$ : as $x \to 2^{+}$ , $\frac{1}{x - 2} \to + \infty$ ; as $x \to 2^{-}$ , it $\to - \infty$ . Why this step? "Every scenario" means also the approach, not just the forbidden point.

$(f \circ g) (x) = \frac{1}{x - 2}, domain x \neq = 2$

Verify: $x = 3 \Rightarrow 1/ (3 - 2) = 1$ ✓. $x = 1.5 \Rightarrow 1/ (- 0.5) = - 2$ ✓ (negative, matching left-approach). $x = 2$ undefined ✓.

Read the figure: the plot of $\frac{1}{x - 2}$ shows the vertical dashed plum asymptote at $x = 2$ . To its right (orange) the curve dives down from $+ \infty$ ; to its left (teal) it climbs from $- \infty$ — exactly the two-sided blow-up described in step 3.

Cell D — Self-composition $f \circ f$

Worked example Example 4 (Cell D)

Given $f (x) = \frac{1}{1 - x}$ . Find $(f \circ f) (x)$ and $(f \circ f \circ f) (x)$ .

Forecast: A weird one — does applying $f$ three times bring you back to $x$ ? Guess yes/no.

Compute $(f \circ f) (x) = f (\frac{1}{1 - x})$ . Substitute $\frac{1}{1 - x}$ into $\frac{1}{1 - □}$ . Take the denominator apart one move at a time: $1 - \frac{1}{1 - x} = \frac{1 - x}{1 - x} - \frac{1}{1 - x} = \frac{( 1 - x ) - 1}{1 - x} = \frac{- x}{1 - x} .$ Why this step? Before we can flip a fraction we must first write its denominator as a single fraction — that means a common denominator $1 - x$ . Now take the reciprocal of that single fraction: $\frac{1}{\frac{- x}{1 - x}} = \frac{1 - x}{- x} = \frac{x - 1}{x} .$ Why this step? Dividing $1$ by a fraction flips it; the sign tidies up to $\frac{x - 1}{x}$ .
Compute the third layer $f ((f \circ f) (x)) = f (\frac{x - 1}{x})$ . Same two moves. First the denominator as one fraction: $1 - \frac{x - 1}{x} = \frac{x}{x} - \frac{x - 1}{x} = \frac{x - ( x - 1 )}{x} = \frac{1}{x} .$ Then flip: $\frac{1}{\frac{1}{x}} = x .$ Why this step? Repeat the identical common-denominator-then-flip routine once more to test for a cycle.

$(f \circ f) (x) = \frac{x - 1}{x} (x \neq = 1, x \neq = 0), (f \circ f \circ f) (x) = x (x \neq = 1, x \neq = 0)$

(Why the restrictions? $f$ itself dies at $x = 1$ ; the second layer feeds $\frac{1}{1 - x}$ into $f$ again, which dies when its input is $1$ , i.e. $x = 0$ . So both $x = 1$ and $x = 0$ are barred throughout — stated here alongside the answer, not hidden below.)

Verify: Take $x = 3$ . $f (3) = \frac{1}{1 - 3} = - \frac{1}{2}$ . $f (- \frac{1}{2}) = \frac{1}{1 - ( - \frac{1}{2} )} = \frac{1}{\frac{3}{2}} = \frac{2}{3}$ . $f (\frac{2}{3}) = \frac{1}{1 - \frac{2}{3}} = \frac{1}{\frac{1}{3}} = 3$ ✓ — three applications return $3$ .

Read the figure: the plum triangle of arrows shows the 3-cycle $3 \to - \frac{1}{2} \to \frac{2}{3} \to 3$ . Each arrow is one application of $f$ ; after exactly three hops you land back on the start — the visual meaning of $(f \circ f \circ f) (x) = x$ .

Cell E — Inverse pair (composition = identity)

Read the figure: $f$ (orange) and $g$ (teal) are drawn as straight lines that are mirror images across the dotted plum $y = x$ line. Reflecting one across $y = x$ lands exactly on the other — the graphical fingerprint of an inverse pair, matching the algebraic collapse to $x$ .

When does an inverse even exist?

only when

f

is bijective — see Bijective functions.

Cell F — Decompose a given function

Read the figure: the onion diagram shows three nested shells — the plum core $x^{2} + 1$ ( $k$ ), the teal shell ( $g$ ), the orange skin $\frac{1}{x}$ ( $f$ ). Peeling from the outside inward is how you read $h$ ; building from the core outward is how you compute it.

Cell G — Real-world word problem (mind the units)

Worked example Example 7 (Cell G)

A factory: hours worked → gadgets → money. Gadgets made from $h$ hours: $g (h) = 12 h$ (gadgets). Revenue from $n$ gadgets: $r (n) = 8 n + 50$ (rupees, the $50$ is a fixed daily bonus). Find revenue as a function of hours, then revenue for a $6$ -hour shift.

Forecast: Which function is inner? (Hours come first in time!)

Identify the chain order. Hours enter first → $g$ is inner; money comes out last → $r$ is outer. So we want $(r \circ g) (h)$ . Why this step? Real processes fix the order: you must make gadgets before you can sell them.
Compose: $(r \circ g) (h) = r (12 h) = 8 (12 h) + 50 = 96 h + 50$ rupees. Why this step? Feed $g$ 's gadget-count output straight into $r$ 's money rule; units flow hours→gadgets→rupees.
Evaluate at $h = 6$ : $96 (6) + 50 = 576 + 50 = 626$ rupees. Why this step? Answer the concrete question with the composite formula.

$(r \circ g) (h) = 96 h + 50, (r \circ g) (6) = 626 rupees$

Verify: Step by step: $g (6) = 72$ gadgets, $r (72) = 8 (72) + 50 = 576 + 50 = 626$ ✓. Units check: (rupees/gadget)(gadgets/hour)(hours)=rupees ✓.

Read the figure: the unit-tracking chain shows three labelled boxes — hours (teal) → gadgets (plum) → rupees (orange) — with the conversion factor written on each arrow. The point $h = 6$ is traced through: $6$ h → $72$ gadgets → $626$ rupees, matching the verify line.

Cell H — Exam twist: solve $f (g (x)) = k$

Read the figure: the parabola $y = x^{2} + 4$ is cut by the horizontal plum line $y = 13$ . The two intersection points sit symmetrically at $x = - 3$ and $x = + 3$ — the picture-proof that a squared inner function hands you a $\pm$ pair.

Cell I — Sign / piecewise & limiting behaviour

The absolute value $∣ x ∣$ is the classic sign-splitter: $∣ x ∣ = x$ when $x \geq 0$ , and $∣ x ∣ = - x$ when $x < 0$ . We compose with it so every sign region appears.

Read the figure: the horizontal axis is the input $x$ , the vertical axis is $(f \circ g) (x) = ∣ x - 2∣$ . The deep-teal line is the left arm ( $x < 2$ , rule $2 - x$ , slope $- 1$ ); the burnt-orange line is the right arm ( $x \geq 2$ , rule $x - 2$ , slope $+ 1$ ); the plum dot marks the corner at $x = 2$ where both arms meet at height $0$ . The two small arrows show both arms rising to $+ \infty$ .

Worked example Example 9 (Cell I)

Given $f (x) = ∣ x ∣$ and $g (x) = x - 2$ . Describe $(f \circ g) (x) = ∣ x - 2∣$ over all sign regions, and its limit as $x \to \pm \infty$ .

Forecast: Where is the "corner"? What does the graph do far left and far right?

Compose: $(f \circ g) (x) = ∣ x - 2∣$ . Why this step? $g$ shifts the input, $f$ folds negatives up — the composite is a shifted V (the figure above).
Case split on the sign of $g (x) = x - 2$ :
- $x \geq 2$ (inner output $\geq 0$ ): $∣ x - 2∣ = x - 2$ , a line of slope $+ 1$ — the orange right arm.
- $x < 2$ (inner output $< 0$ ): $∣ x - 2∣ = - (x - 2) = 2 - x$ , a line of slope $- 1$ — the teal left arm. Why this step? Absolute value has two rules; we must show both sign regions.
Corner and limits: the two pieces meet at $x = 2$ where value $= 0$ (the plum vertex). As $x \to + \infty$ , $∣ x - 2∣ \to + \infty$ ; as $x \to - \infty$ , $∣ x - 2∣ \to + \infty$ . Why this step? Degenerate/limiting behaviour completes the picture — the V opens upward forever on both sides.

$(f \circ g) (x) = ∣ x - 2∣ = {x - 2, 2 - x, x \geq 2 x < 2$

Verify: $x = 5 \Rightarrow ∣3∣ = 3$ and $5 - 2 = 3$ ✓ (right piece). $x = 0 \Rightarrow ∣ - 2∣ = 2$ and $2 - 0 = 2$ ✓ (left piece). $x = 2 \Rightarrow ∣0∣ = 0$ ✓ (vertex). All three land on the V drawn above.

Recall Self-test before you close the tab

Compute $(g \circ f) (3)$ for Ex 1's functions. ::: $g (f (3)) = g (5) = 25$ . Domain of $5 - x$ ? ::: $(- \infty, 5]$ . What signals that $f$ and $g$ are inverses? ::: both compositions equal $x$ (the identity). Where does $\frac{1}{x - 2}$ blow up? ::: at $x = 2$ . Innermost function of $\frac{1}{x ^{2} + 1}$ ? ::: $k (x) = x^{2} + 1$ .