MA4 Functions

Lesson

A function is a rule with one job: take an input, give back exactly one output. Almost everything the ACT asks about functions is a test of whether you can read that rule as an instruction rather than as decoration. f(3) does not mean f times 3. It means: wherever you see x in the rule, write 3, then work it out.

Read the notation out loud in words and most of these questions stop being tricky. f(x) = 2x + 1 says "double the input and add one". So f(3) is "double 3 and add one" — 7. f(a + 4) is "double a + 4 and add one" — 2(a + 4) + 1, and the parentheses matter, because the whole expression a + 4 is the input.

The mistake that costs the most points

For f(g(4)), the overwhelmingly common error is to compute g(4) and stop, because that is where the work starts to feel finished. It is not finished. g(4) is only the number you hand to f. Work strictly from the inside out, and write the intermediate number down where you can see it:

• Compute the inside function at the given input. Circle the result.
• Feed that number — not the original input — into the outside function.
• Check the order the question asked for. f(g(x)) and g(f(x)) are usually different functions, and the ACT offers both.

The same habit fixes inverses. f⁻¹ is not 1/f. The −1 is notation for "the function that undoes f", so f⁻¹(11) asks: what input would f have turned into 11? That means setting the rule equal to 11 and solving, which is why the method is called swap-and-solve — write y = rule, swap x and y, solve for y. And undo the operations in reverse order: if f multiplies and then adds, f⁻¹ subtracts and then divides.

Transformations run backwards from your intuition — inside only

Everything outside the function behaves the way you expect: f(x) + 3 moves the graph up 3, −f(x) flips it across the x-axis, 2f(x) stretches every height to twice what it was. Everything inside the parentheses acts on the input and runs the other way.

• f(x + 3) moves the graph LEFT 3, not right. Ask which input now produces the old output: to make the inside read 0, x must be −3. The graph reaches each value three units earlier.
• f(x − 3) moves it right 3, for the same reason in reverse.
• f(−x) reflects across the y-axis; −f(x) reflects across the x-axis. Inside means left–right, outside means up–down.

If the direction ever feels uncertain, test one point. Take a point you know is on y = f(x), say (2, 5), and ask what x makes the new inside equal 2. For y = f(x − 4) + 3 you need x = 6, and the height becomes 8 — so the point moved right and up, and you never had to trust a rule you half-remember.

Getting the vertex out of whichever form you are handed

The three forms of a quadratic are the same function wearing different clothes. Each one displays a different feature for free, and the ACT chooses the form that hides what it is asking for.

• Vertex form, a(x − h)² + k: the vertex is (h, k) — read it off. Watch the sign: in (x − 4)² the vertex is at x = +4. If a > 0 the parabola opens upward and k is the minimum; if a < 0 it opens downward and k is the maximum.
• Factored form, a(x − p)(x − q): the x-intercepts are p and q. The axis of symmetry is exactly halfway between them, at their average, and the vertex sits on that line.
• Standard form, ax² + bx + c: c is the y-intercept, and the axis of symmetry is x = −b/(2a). That formula gives you the x-coordinate only — you still have to substitute it back into the function to get how high the vertex is. Half of all vertex errors are stopping one step early there.

Use the answer choices — they are four extra pieces of information

Every ACT math question is multiple choice with four options, and there is no penalty for a wrong guess, so an answer you can verify is worth more than an answer you can derive elegantly. Substituting the choices back into the question is often the fastest legitimate route:

• Inverse and solving questions: testing whether f(9) = 39 is one line of arithmetic; solving from scratch is three.
• "Which expression is equivalent to f(g(x))?": pick a convenient input like x = 2, run it through the real composition, then evaluate each choice at x = 2. Only the equivalent one matches — and this catches expansion slips no amount of care prevents.
• Sequences: if a choice for the 20th term is smaller than the 4th term of an increasing sequence, it is gone before you compute anything.

One caution about testing choices: an answer that works for a single input is not always equivalent. Try x = 2 and also x = −1 if two expressions both survive. And do not choose questions to skip by their position — ACT does not publish any claim that its questions run from easiest to hardest, and its own advice is to answer the ones you find quick first, then come back.

Growth, decay and the sequences that hide inside them

Repeated multiplying is exponential: a quantity that doubles every 4 hours is multiplied by 2 once per period, so in 20 hours it has doubled 20 ÷ 4 = 5 times and the multiplier is 2⁵, not 2 × 5. Halving works identically with a factor of ½. A percent change becomes a single multiplier: down 15% each year means multiply by 0.85 each year, giving V(t) = V₀(0.85)ᵗ.

• Arithmetic sequence (constant difference): a(n) = a(1) + (n − 1)d. The n − 1 is where points are lost — getting to the 20th term takes 19 steps, not 20.
• Sum of an arithmetic sequence: number of terms × the average of the first and last term.
• Geometric sequence (constant ratio): a(n) = a(1)·r^(n−1) — again one less than the term number.

Piecewise functions ask the same first question as everything else: which rule applies to this input? Check the conditions before you substitute, take special care at the boundary value, where the inequality signs decide which line owns it, and then use only that line.