4.2 Functions & Function Notation

What Is a Function?

A function is a rule that assigns to each input exactly one output. Think of it as a machine: put in a number x, get out a single number y. The key restriction is the word exactly one — an input is never allowed to produce two different outputs.

Functions can be given as equations (y = 2x + 1), tables, graphs, or sets of ordered pairs. The set {(1, 3), (2, 5), (3, 7)} is a function because each first coordinate appears once. But {(1, 3), (1, 4)} is not a function, because the input 1 maps to both 3 and 4.

The vertical line test. A graph represents a function if and only if no vertical line crosses it more than once. If any vertical line hits the graph twice, that x-value has two y-values, violating the definition. A straight line (non-vertical) passes the test; a parabola y = x² passes; but a sideways parabola x = y² or a full circle fails — a vertical line through it hits two points.

Worked example. Is the circle x² + y² = 25 a function? Solving for y gives y = ±√(25 − x²). The ± means most x-values (like x = 0 → y = ±5) produce two outputs, so a vertical line cuts it twice. It is not a function.

Domain, Range & Function Notation

The domain is the set of all allowed input (x) values; the range is the set of all resulting output (y) values. On a placement test, the two domain restrictions to watch for are:

• Division by zero — exclude any x that makes a denominator 0.
• Even roots of negatives — the inside of a square root must be ≥ 0.

Worked example (domain). Find the domain of f(x) = 1/(x − 4). The denominator is 0 when x = 4, so the domain is all real numbers except x = 4.

Worked example (domain with a root). Find the domain of g(x) = √(x − 3). The radicand must satisfy x − 3 ≥ 0, so x ≥ 3. The domain is [3, ∞).

Function notation writes the output as f(x), read "f of x." It does not mean f times x — it names the output of function f at input x. To evaluate, substitute the value everywhere x appears.

Worked example. If f(x) = 3x² − 2x + 1, find f(−2):

• f(−2) = 3(−2)² − 2(−2) + 1 = 3(4) + 4 + 1 = 12 + 4 + 1 = 17.

Notice (−2)² = 4 (square first), and −2(−2) = +4. These sign steps are where errors creep in.

Composite Functions & Transformations

A composite function chains two functions: (f∘g)(x) = f(g(x)) means "do g first, then apply f to that result." Work from the inside out.

Worked example. Let f(x) = x + 5 and g(x) = x². Find (f∘g)(3) and (g∘f)(3).

• (f∘g)(3) = f(g(3)) = f(9) = 9 + 5 = 14.
• (g∘f)(3) = g(f(3)) = g(8) = 8² = 64.

Order matters — composition is not commutative.

Basic transformations shift or reflect a graph y = f(x):

Worked example. Starting from y = x², the graph y = (x − 3)² + 2 is the parabola shifted 3 right and 2 up, so its vertex moves from (0, 0) to (3, 2).

Common mistake: Horizontal shifts feel backwards — f(x + 4) moves the graph left 4, not right. The shift is opposite the sign inside the parentheses. Also, do not read f(x) as multiplication: f(x) = x² means "square the input," so f(3) = 9, never 3x.

Reading Functions from Graphs & Tables

A placement test often shows a graph and asks you to read off values. To find f(a), locate x = a on the horizontal axis, move up or down to the curve, then read the y-value. To solve f(x) = b, do the reverse: find where the height equals b and read the x-value(s).

Worked example. Suppose a graph passes through (0, 4), (2, 0), and (4, −4). Then f(0) = 4, f(2) = 0, and f(4) = −4. The point where f(x) = 0 — here x = 2 — is called a zero or root of the function, the x-intercept of its graph.

Range from a graph. The range is the set of all y-values the graph actually reaches. A parabola y = x² opening upward from vertex (0, 0) has range y ≥ 0, because outputs are never negative — squaring can't produce a negative number.

Worked example (evaluate a piecewise idea). If f(x) = |x| (absolute value), then f(−5) = 5 and f(5) = 5; the V-shaped graph has its lowest point at the origin and range y ≥ 0.

Even vs. odd quick check. A graph symmetric about the y-axis satisfies f(−x) = f(x) (even, like x²); one symmetric about the origin satisfies f(−x) = −f(x) (odd, like x³). You won't need formal proofs on a placement test, but recognizing this symmetry helps you sketch and check answers quickly. Always confirm a candidate output by re-substituting it into the rule.