6.3 Functions, Notation & Transformations

Function Notation and Evaluation

Official Guide algebra (OG 3.2) includes functions. On Problem Solving, a function is a rule that sends each allowed input to exactly one output. The notation y = f(x) means "the output of rule f at input x." The set of legal inputs is the domain; the outputs that actually occur are the range.

Evaluating f(argument) means replacing every x with the entire argument, wrapped in parentheses.

Numerical. If f(x) = 3x² − 5x + 2, then f(4) = 3(16) − 5(4) + 2 = 48 − 20 + 2 = 30.

Algebraic. f(x + 2) = 3(x + 2)² − 5(x + 2) + 2 = 3(x² + 4x + 4) − 5x − 10 + 2 = 3x² + 12x + 12 − 5x − 8 = 3x² + 7x + 4.

Reciprocal input. f(1/x) = 3(1/x)² − 5(1/x) + 2 = 3/x² − 5/x + 2.

Trap: f(a + b) is not f(a) + f(b) in general, and f(kx) is not k · f(x) unless f happens to be a linear map through the origin. Expand; do not distribute the letter f.

A graph represents a function of x when it passes the vertical-line test: no vertical line hits the graph more than once. The vertical line x = k from Section 6.2 is itself not a function of x. A horizontal line y = c is a constant function.

Domain restrictions you actually need

On Quant, "all real numbers" is the default universe. Two restrictions appear over and over:

1. Denominator ≠ 0. If f(x) = (2x + 1)/(x² − 9), then x ≠ 3 and x ≠ −3.
2. Even-root radicand ≥ 0. If f(x) = √(16 − 4x), then 16 − 4x ≥ 0 → x ≤ 4.

You will not be asked for calculus-style range proofs. You may be asked for the minimum of a quadratic written as a(x − h)² + k: if a > 0 the minimum output is k, so the range is y ≥ k.

Defined Operations (Invented Symbols)

GMAC likes to define a new binary symbol on the spot. The stem looks like:

For all integers a and b, a ★ b = a² − 2ab + b.

That is just a function of two variables. Substitute in the given order, inner parentheses first. Do not assume the operation is commutative (a ★ b need not equal b ★ a) or associative ((a ★ b) ★ c need not equal a ★ (b ★ c)).

Worked example. a ★ b = a² − 2ab + b. Compute (3 ★ 2) ★ 4.

• Inner: 3 ★ 2 = 9 − 2(3)(2) + 2 = 9 − 12 + 2 = −1
• Outer: (−1) ★ 4 = (−1)² − 2(−1)(4) + 4 = 1 + 8 + 4 = 13

Trap arithmetic: treating (−1)² as −1 produces −1 + 8 + 4 = 11. Dropping the sign in −2ab produces 1 − 8 + 4 = −3. Square the first operand, including a negative base, then handle the middle term.

A unary box is the same idea with one input: ♣x = x² − 3x. Then ♣(♣2) = ♣(4 − 6) = ♣(−2) = 4 − 3(−2) = 4 + 6 = 10. Nested defined operations are composition in costume.

When the definition has a denominator, the inner value cannot make that denominator 0 — the same domain rule as any rational function.

Typical GMAT operator checklist:

• Copy the template with blanks: first operand everywhere the definition uses the first letter, second operand everywhere it uses the second letter.
• If the stem writes (p ★ q) ★ r, compute p ★ q as a single number, then use that number as the new first operand with r.
• If two remaining answers match your inner value, you stopped one layer too soon.
• If the stem asks for a ★ b + b ★ a, compute both orders separately; they often differ.

Light Composition

A composite (f ∘ g)(x) = f(g(x)) is inside-out:

1. Compute u = g(x).
2. Compute f(u).

Order matters: f(g(x)) is not g(f(x)) in general.

Worked example (order). f(x) = x² − 4 and g(x) = 2x + 3.

• f(g(2)): inner g(2) = 7, outer f(7) = 49 − 4 = 45
• g(f(2)): inner f(2) = 0, outer g(0) = 3

Same ingredients, two different outputs. The engine is not being cute; it is testing whether you respect parentheses.

Worked example (solve a composite). f(x) = 2x² − 5, g(x) = 3x + 1, and k is a positive constant with f(g(k)) = 67.

• 2[g(k)]² − 5 = 67 → 2[g(k)]² = 72 → [g(k)]² = 36 → g(k) = 6 or g(k) = −6
• 3k + 1 = 6 → k = 5/3
• 3k + 1 = −6 → k = −7/3, discarded because k is positive

So k = 5/3. The value −7/3 is the extra algebraic root. The value 6 is g(k), not k — a classic stop-one-layer-too-soon trap sitting next to 5/3 in a five-choice live item.

From a table, composition is a lookup chain. If a table gives g(2) = 5 and another row gives f(5) = 11, then f(g(2)) = 11. If f(5) is missing, the composition is not determined from the table. Never average table outputs or "combine" rows.

Shifts: f(x − h) + k

Official QR includes interpreting graphs. The transformation you must read at sight is the rigid shift of a graph:

g(x) = f(x − h) + k

The horizontal rule is the one testers reverse. f(x − 3) shifts right by 3, because the input that reproduces f(0) is now x = 3. Inside the parentheses, the sign looks backwards. Vertical shifts do what they look like: +k is up.

Track a landmark (vertex, intercept, or a plotted point) in a fixed order:

1. Horizontal shift from the inside of f(·)
2. Reflection −f if present
3. Vertical shift ± k

Worked example. Vertex of y = f(x) is (3, −4). Let g(x) = −f(x + 2) + 3.

• f(x + 2) moves left 2: x-coordinate 3 − 2 = 1
• −f reflects over the x-axis: y-coordinate −(−4) = 4
• +3 moves up 3: y-coordinate 4 + 3 = 7
• New vertex (1, 7)

If you shift (3, −4) right because you saw "+2," you land on x = 5 and miss. If you forget the reflection and add 3 to −4, you land on y = −1, another designed neighbor.

Reading a graphed line or curve

When the item shows a graph rather than an equation:

• A straight graph is a line: read two grid points, compute slope, read the y-intercept off the axis (Section 6.2).
• A V-shape is typically y = a|x − h| + k: vertex (h, k), opening up if a > 0.
• A shift of y = x² has vertex (h, k) from y = (x − h)² + k, which is exactly f(x − h) + k with f(x) = x².

If the graph of f is given and you need g(x) = f(x) + 2, every labeled y-coordinate increases by 2. If you need f(x − 1), every labeled point slides right 1; the y-value at a fixed x generally changes because you are reading a different input.

Worked example (graph reading). A plotted point on y = f(x) is (2, 5). On y = f(x − 4) + 1 that point moves to (2 + 4, 5 + 1) = (6, 6). The output f(2) is still 5; you have moved the point, not substituted x = 2 into the new rule blindly without shifting.

Keep function evaluation and graph shifts in separate mental boxes when the stem mixes them. f(2) is a number. The graph of y = f(x − 2) is a picture moved right. Computing f(2) does not tell you where the moved graph crosses the y-axis unless you do the shift on an actual known point.