Function notation — f(x), g(x): Worked examples

The scenario matrix

Every function-notation question is one of these "case classes". A blank cell means "we still owe you an example there" — by the end, none are blank.

Case class	What makes it tricky	Covered in
Positive input	plain substitution	Ex 1
Negative input	sign traps: $(- a)^{2} = + a^{2}$	Ex 1
Zero input	most terms vanish → the $y$ -intercept	Ex 1
Fraction / decimal input	arithmetic care, no new rule	Ex 2
Expression as input ( $f (a + h)$ )	replace every $x$ by the whole blob	Ex 3
Composition $f (g (x))$ vs $g (f (x))$	order matters, inside-out	Ex 4
Solving $f (x) = c$ (working backwards)	notation read in reverse	Ex 5
Difference quotient (limiting behaviour)	the gateway to Limits and calculus	Ex 6
Word problem (real units)	translate English → machine	Ex 7
Exam twist ( $f (f (x))$ , piecewise)	self-composition + Piecewise functions	Ex 8

Let me set two functions we will reuse: $f (x) = 3 x^{2} - 2 x + 5, g (x) = 2 x - 3.$

Worked examples

Worked example Example 1 — Positive, negative AND zero input (cells: positive / negative / zero)

For $f (x) = 3 x^{2} - 2 x + 5$ , find $f (2)$ , $f (- 1)$ and $f (0)$ .

Forecast: Guess the sign of each answer. Will $f (- 1)$ be bigger or smaller than $f (2)$ ? (Hint: negatives sometimes make quadratics bigger, not smaller.)

Step 1 — $f (2)$ : substitute $2$ for every $x$ . $f (2) = 3 (2)^{2} - 2 (2) + 5 = 3 \cdot 4 - 4 + 5 = 13.$ Why this step? The rule $f$ says "square it, times 3, minus twice it, plus 5". Every $x$ is a slot; we drop $2$ into all of them.

Step 2 — $f (- 1)$ : the sign trap. $f (- 1) = 3 (- 1)^{2} - 2 (- 1) + 5.$ Why this step? Two danger points: $(- 1)^{2} = + 1$ (a square is never negative), and $- 2 \times (- 1) = + 2$ (minus times minus is plus). Both terms turn positive. $= 3 (1) + 2 + 5 = 10.$

Step 3 — $f (0)$ : watch the terms disappear. $f (0) = 3 (0)^{2} - 2 (0) + 5 = 0 - 0 + 5 = 5.$ Why this step? Anything times $0$ is $0$ , so only the constant survives. That leftover number is exactly where the graph crosses the vertical axis — the $y$ -intercept. See the picture.

Verify: Plug the outputs back as points. The parabola through $(2, 13)$ , $(- 1, 10)$ , $(0, 5)$ opens upward and its lowest point sits between $x = - 1$ and $x = 2$ — consistent with all three heights being positive. ✓

Worked example Example 3 — A whole expression as input (cell: expression input)

For $f (x) = 3 x^{2} - 2 x + 5$ , find $f (a + h)$ , fully expanded.

Forecast: Will the answer contain a lone $h^{2}$ term? Yes/No — decide before reading.

Step 1 — Replace every $x$ by the blob $(a + h)$ . $f (a + h) = 3 (a + h)^{2} - 2 (a + h) + 5.$ Why this step? The slot doesn't know if you feed it a number or an expression. The whole $(a + h)$ is one input — brackets keep it together so we don't square only part of it.

Step 2 — Expand $(a + h)^{2}$ . $(a + h)^{2} = a^{2} + 2 ah + h^{2} .$ Why this step? This is $(a + h) (a + h)$ multiplied out; forgetting the middle $2 ah$ is the classic slip.

Step 3 — Distribute and collect. $f (a + h) = 3 a^{2} + 6 ah + 3 h^{2} - 2 a - 2 h + 5.$ Why this step? Multiply the $3$ and $- 2$ through, then there's nothing to combine — every term has a different mix of $a$ and $h$ .

Verify: Set $h = 0$ : it must collapse back to $f (a)$ . Indeed $3 a^{2} + 0 + 0 - 2 a - 0 + 5 = 3 a^{2} - 2 a + 5 = f (a)$ . ✓ (And yes, the $3 h^{2}$ term survives.)

Worked example Example 4 — Order matters:

$f (g (x))$ vs $g (f (x))$ (cell: composition) Using $f (x) = 3 x^{2} - 2 x + 5$ and $g (x) = 2 x - 3$ , compute $f (g (2))$ and $g (f (2))$ .

Forecast: Will these two be equal? (Recall composition is generally not commutative — see Composite functions.)

Step 1 — $f (g (2))$ : inside first. $g (2) = 2 (2) - 3 = 1.$ Why this step? Read inside-out. You cannot feed $f$ until you know the value coming out of $g$ .

Step 2 — Feed that into $f$ . $f (1) = 3 (1)^{2} - 2 (1) + 5 = 3 - 2 + 5 = 6.$ Why this step? Now $1$ is the input to $f$ .

Step 3 — $g (f (2))$ : swap the order. $f (2) = 13 (from Ex 1), g (13) = 2 (13) - 3 = 23.$ Why this step? Now $f$ runs first. Its output $13$ becomes $g$ 's input.

Verify: $6 \neq = 23$ , confirming order matters. Sanity: $g$ (a line) is gentle, $f$ (a parabola) blows up — so whichever machine runs last dominates the size, and $f$ -last ( $= 6$ ) vs $g$ -last ( $= 23$ )... wait, $f$ -last gave the smaller answer because its input $1$ was small. The lesson: the inner value drives everything. ✓

Worked example Example 6 — The difference quotient (cell: limiting behaviour)

For $f (x) = 3 x^{2} - 2 x + 5$ , simplify the difference quotient $\frac{f ( x + h ) - f ( x )}{h},$ then say what happens as $h$ shrinks toward $0$ .

Forecast: After cancelling, will $h$ disappear completely, or leave a leftover $h$ term?

Step 1 — Build $f (x + h)$ (same trick as Ex 3). $f (x + h) = 3 x^{2} + 6 x h + 3 h^{2} - 2 x - 2 h + 5.$ Why this step? The difference quotient measures how much output changes between input $x$ and a nearby input $x + h$ — we need both.

Step 2 — Subtract $f (x)$ . $f (x + h) - f (x) = 6 x h + 3 h^{2} - 2 h .$ Why this step? The parts with no $h$ ( $3 x^{2}$ , $- 2 x$ , $+ 5$ ) are identical in both and cancel. Only the change remains, and notice every surviving term has a factor of $h$ .

Step 3 — Divide by $h$ . $\frac{6 x h + 3 h ^{2} - 2 h}{h} = 6 x + 3 h - 2.$ Why this step? Because every term had an $h$ , division is clean and legal (we're not dividing by zero yet — $h$ is still nonzero here).

Step 4 — Let $h \to 0$ . $6 x + 3 h - 2 ⟶ 6 x - 2.$ Why this step? As the second input slides onto the first, the leftover $3 h$ vanishes. What's left, $6 x - 2$ , is the slope of $f$ at $x$ — the doorway to Limits and calculus.

Verify: At $x = 1$ the slope is $6 (1) - 2 = 4$ . Check with a tiny $h = 0.001$ : $\frac{f ( 1.001 ) - f ( 1 )}{0.001} = 6 (1) + 3 (0.001) - 2 = 4.003 \approx 4$ . ✓

Worked example Example 7 — Word problem with real units (cell: word problem)

A phone plan costs $\$ 8 $ba se pl u s$ $0.05 $p er min u t e . W r i t e i t a s$ C(m) $, t h e n f in d$ C(200) $an d t h e min u t es t ha t mak e t h e bi l l$ $20$.

Forecast: Will $200$ minutes cost more or less than $\$ 20$? Guess before computing.

Step 1 — Translate English into a machine. $C (m) = 8 + 0.05 m,$ where $m$ is minutes and $C (m)$ is dollars. Why this step? "Base plus per-minute" is a starting value plus a rate times the input — exactly a linear rule. Naming it $C$ (for cost) with input $m$ makes units explicit.

Step 2 — Evaluate $C (200)$ . $C (200) = 8 + 0.05 (200) = 8 + 10 = 18 dollars .$ Why this step? Substitute $200$ minutes; the $0.05 \cdot 200$ has units $\tfrac{\$ }{\text{min}}\times\text{min} = $$, so it's dollars — units check.

Step 3 — Solve $C (m) = 20$ (backwards, like Ex 5). $8 + 0.05 m = 20 \Rightarrow 0.05 m = 12 \Rightarrow m = 240 min .$ Why this step? We want the input that produces a $\$ 20$ output — read the notation in reverse.

Verify: $C (240) = 8 + 0.05 (240) = 8 + 12 = 20$ . ✓ And $200$ min cost only $\$ 18 < $20$, matching a correct forecast.

Worked example Example 8 — Exam twist: self-composition of a piecewise machine (cell: exam twist)

Let $h (x) = {x + 4, x^{2}, x < 0 x \geq 0$ a piecewise function. Find $h (h (- 3))$ .

Forecast: The inner value $h (- 3)$ — will it land in the $x < 0$ branch or the $x \geq 0$ branch? That choice decides everything.

Step 1 — Inner: $h (- 3)$ , choose the right branch. Since $- 3 < 0$ , use $x + 4$ : $h (- 3) = - 3 + 4 = 1.$ Why this step? A piecewise machine first checks which zone the input is in, then applies that zone's rule. $- 3$ is negative → top branch.

Step 2 — Outer: $h (1)$ , re-check the branch. Now the input is $1$ , and $1 \geq 0$ , so use $x^{2}$ : $h (1) = 1^{2} = 1.$ Why this step? The output $1$ from step 1 is a new input that must be re-tested against the conditions. It jumped from the negative branch to the non-negative branch — a common trap.

Verify: $h (h (- 3)) = h (1) = 1$ . Cross-check by testing the branch boundaries near $x = 0$ : at $x = 0$ , $x^{2} = 0$ ; just below, $x + 4 \approx 4$ — the machine genuinely behaves differently on each side, so branch-checking mattered. ✓

Recall Rapid self-test

$f (0)$ for $f (x) = 3 x^{2} - 2 x + 5$ ::: $5$ $f (- 1)$ for the same $f$ ::: $10$ Does $f (g (2)) = g (f (2))$ for $f, g$ above? ::: No — $6 \neq = 23$ , composition isn't commutative The $h \to 0$ limit of the difference quotient of $f$ ::: $6 x - 2$ $h (h (- 3))$ for the piecewise $h$ ::: $1$ Minutes so the phone bill is $\$ 20 $:::$ 240$

Connections

Function notation — f(x), g(x) — the parent this page drills.
Composite functions — Examples 4 and 8 are composition in action.
Inverse functions — Examples 5 and 7 solve $f (x) = c$ , the reverse machine.
Piecewise functions — Example 8's branch-checking.
Limits and calculus — Example 6's difference quotient is the first step of the derivative.
Graphing functions — Example 1's zero input is the $y$ -intercept.