Function notation — f(x), g(x): Exercises — Maths deep dive

Before we start, one picture to keep in your head: a function is a machine with a name on the side and a rule inside. You drop an input in the top; a single output falls out the bottom.

Level 1 — Recognition

Goal: read the notation and substitute a single number correctly.

Recall Solution 1.1

The rule reads: "multiply the input by 4, then subtract 7." We replace every $x$ with the given number.

What we did / why: substitution is the whole job of $f (something)$ — put the something wherever $x$ sits. $f (3) = 4 (3) - 7 = 12 - 7 = 5$ $f (0) = 4 (0) - 7 = 0 - 7 = - 7$ $f (- 2) = 4 (- 2) - 7 = - 8 - 7 = - 15$ Notice $f (0) = - 7$ is the y-intercept — the output when the input is zero.

Recall Solution 1.2

Here the input letter is $t$ , but nothing changes — the letter is just a placeholder. $g (4) = 4^{2} - 5 = 16 - 5 = 11$ $g (- 4) = (- 4)^{2} - 5 = 16 - 5 = 11$ What it looks like: $(- 4)^{2}$ is $(- 4) \times (- 4) = + 16$ , not $- 16$ . Because squaring erases the sign, $g (4) = g (- 4)$ . The graph is a mirror across the vertical axis — this function is even. See Graphing functions.

Level 2 — Application

Goal: feed a whole expression into a function and simplify.

Recall Solution 2.1

The input is now an expression. We replace every $x$ with the entire thing, brackets and all.

$f (x + 1) = (x + 1)^{2} + 3$ Why brackets matter: the $+ 3$ is outside; only $(x + 1)$ is squared. Expanding $(x + 1)^{2} = x^{2} + 2 x + 1$ : $f (x + 1) = x^{2} + 2 x + 1 + 3 = x^{2} + 2 x + 4$

$f (2 x) = (2 x)^{2} + 3 = 4 x^{2} + 3$ What it looks like: $(2 x)^{2} = 2^{2} \cdot x^{2} = 4 x^{2}$ — the whole $2 x$ gets squared, not just the $x$ .

Recall Solution 2.2

$h (x + 2) = 2 (x + 2) - 1 = 2 x + 4 - 1 = 2 x + 3$ Here the input itself was shifted, so the whole rule acts on $x + 2$ . $h (x) + 2 = (2 x - 1) + 2 = 2 x + 1$ Here we ran the machine on $x$ , then added 2 to the output. $2 x + 3 \neq = 2 x + 1$ They differ by exactly $2$ — the mistake would cost you a constant every time.

Level 3 — Analysis

Goal: work inside-out through composition and read differences.

Recall Solution 3.1

Rule: work from the inside out — the function nearest the number acts first.

$f (g (2))$ : first $g (2) = 2^{2} = 4$ , then $f (4) = 3 (4) + 2 = 14$ . $g (f (2))$ : first $f (2) = 3 (2) + 2 = 8$ , then $g (8) = 8^{2} = 64$ . $f (g (2)) = 14, g (f (2)) = 64, 14 \neq = 64$ Order matters — composition is not commutative. More in Composite functions.

Recall Solution 3.2

Why this expression? It measures the average steepness of $f$ over a small step $h$ — the seed of the derivative in Limits and calculus.

Numerator, piece by piece: $f (x + h) = (x + h)^{2} = x^{2} + 2 x h + h^{2}$ $f (x + h) - f (x) = (x^{2} + 2 x h + h^{2}) - x^{2} = 2 x h + h^{2}$ Now divide by $h$ . Since $h \neq = 0$ we may cancel it: $\frac{2 x h + h ^{2}}{h} = \frac{h ( 2 x + h )}{h} = 2 x + h$ What it looks like: as the step $h$ shrinks toward $0$ , this approaches $2 x$ — the slope at the point.

Level 4 — Synthesis

Goal: solve for inputs, invert, and combine several ideas.

Recall Solution 4.1

What we're asked: run the machine backwards — which input produces output 21? Set the rule equal to 21 and solve: $5 x - 4 = 21$ Add 4 to both sides (undo the subtraction): $5 x = 25$ . Divide by 5 (undo the multiplication): $x = 5$ . $x = 5$ Check forward: $f (5) = 5 (5) - 4 = 21$ . ✓ This "undoing" is exactly what Inverse functions formalises.

Recall Solution 4.2

$f (g (x)) = \frac{( 2 x - 3 ) + 3}{2} = \frac{2 x}{2} = x$ $g (f (x)) = 2 (\frac{x + 3}{2}) - 3 = (x + 3) - 3 = x$ What it means: each function undoes the other — both compositions collapse to $x$ . So $f$ and $g$ are inverses of one another. Feeding an input through one then the other returns you to where you started.

Recall Solution 4.3

Step 1 — pick the correct piece by checking the condition on the input.

$- 3 < 0$ , use $x^{2}$ : $p (- 3) = (- 3)^{2} = 9$ .
$0 \geq 0$ (note the $\geq$ , so $0$ takes the second branch): $p (0) = 2 (0) + 1 = 1$ .
$4 \geq 0$ , use $2 x + 1$ : $p (4) = 2 (4) + 1 = 9$ . $p (- 3) = 9, p (0) = 1, p (4) = 9$ The boundary point $x = 0$ is decided entirely by which side gets the " $=$ ". See Piecewise functions.

Level 5 — Mastery

Goal: chain everything — algebra, composition, inverses, generality.

Recall Solution 5.1

Translate the facts into equations by substituting: $f (1) = a (1) + b = a + b = 5$ $f (3) = a (3) + b = 3 a + b = 11$ Subtract the first from the second (the $b$ 's cancel): $(3 a + b) - (a + b) = 11 - 5 \Rightarrow 2 a = 6 \Rightarrow a = 3$ Back-substitute into $a + b = 5$ : $3 + b = 5 \Rightarrow b = 2$ . So $f (x) = 3 x + 2$ . $f (10) = 3 (10) + 2 = 32$

Recall Solution 5.2

Build the composite first (inside-out): $f (g (x)) = (x - 4)^{2} + 1$ Set equal to 10: $(x - 4)^{2} + 1 = 10 \Rightarrow (x - 4)^{2} = 9$ Cover both cases — a square equals 9 when its base is $+ 3$ or $- 3$ : $x - 4 = 3 \Rightarrow x = 7$ $x - 4 = - 3 \Rightarrow x = 1$ $x = 7 or x = 1$ Check: $g (7) = 3, f (3) = 10$ ✓ and $g (1) = - 3, f (- 3) = 10$ ✓. Forgetting the negative root loses a whole solution.

Recall Solution 5.3

Feed $f$ into itself: $f (f (x)) = f (\frac{1}{x}) = \frac{1}{1/ x} = x$ because dividing 1 by $1/ x$ flips the fraction back. So $f$ is its own inverse (for $x \neq = 0$ ). Degenerate case: at $x = 0$ the inner $f (x) = 1/ x$ is undefined — you cannot even start, so $x = 0$ is excluded from the domain. As $x \to 0^{+}$ , $f (x) \to + \infty$ ; as $x \to 0^{-}$ , $f (x) \to - \infty$ . The output shoots off in opposite directions on the two sides. See Limits and calculus.

Recall Self-test checklist

Read the notation without substituting numbers ::: L1 — done if 1.1, 1.2 were instant Substitute a whole expression and keep brackets ::: L2 — 2.1, 2.2 Work composition inside-out ::: L3 — 3.1 Build and cancel a difference quotient ::: L3 — 3.2 Invert a function / recognise inverse pairs ::: L4 — 4.1, 4.2, 5.3 Choose the right piecewise branch at a boundary ::: L4 — 4.3 Solve for unknown constants and both roots ::: L5 — 5.1, 5.2

Connections

Function notation — f(x), g(x) — the parent note these exercises drill
Composite functions — inside-out evaluation (Ex 3.1, 4.2, 5.2, 5.3)
Inverse functions — undoing a machine (Ex 4.1, 4.2, 5.3)
Piecewise functions — branch selection (Ex 4.3)
Function transformations — input-shift vs output-shift (Ex 2.2)
Limits and calculus — difference quotient, behaviour near a hole (Ex 3.2, 5.3)
Graphing functions — even symmetry (Ex 1.2)