f(x) is a name with an instruction inside it, not a multiplication.
Work out the fare
A tricycle charges ₱25 for the first 2 km and ₱8 for each kilometre after that. What is the fare for a 6 km ride?
The first 2 km cost ₱25. That leaves 6 − 2 = 4 km at ₱8, which is ₱32. The fare is 25 + 32 = ₱57. D charges all 6 km at ₱8 on top of the ₱25 — but the ₱25 already paid for the first two. B forgets the ₱25 entirely, and A charges only one extra kilometre. Written as a function, f(x) = 25 + 8(x − 2) for x ≥ 2, and f(6) = 57.
One input, one output. Two inputs may share an output — that is allowed.
On a graph the same rule is the vertical line test: if any vertical line crosses the graph twice, that x has two y values and the graph is not a function.
The brackets are part of the name, the way they are in sin(30°). Writing f(2) = 2f and then "cancelling" the f is the single most common slip in this topic, and it produces answers that are not even in the right units.
f(x) = x² − 4x + 3, find f(−2)
the rule and the input
f(−2) = (−2)² − 4(−2) + 3
brackets around the −2
= 4 + 8 + 3
a negative squared is positive
= 15
f(−2) = 15
Write the brackets. Without them −2² reads as −(2²) = −4, and the whole answer shifts by 8.
f(x) = 2x + 5, find f(a + 1)
a + 1 goes in wherever x was
f(a + 1) = 2(a + 1) + 5
brackets again
= 2a + 2 + 5 = 2a + 7
expand and tidy
Check it against a number you can do both ways. At a = 3, f(4) = 13, and 2(3) + 7 = 13 ✓ This check costs ten seconds and catches a wrong expansion every time.
Some questions give the output and ask for the input. Set the rule equal to that output and solve.
Given f(x) = 4x − 9 and f(k) = 15, find k.
4k − 9 = 15
the rule, with k in, equals the output
4k = 24
add 9
k = 6
divide by 4
f(6) = 24 − 9 = 15 ✓
check in the rule
k is an input, so the check is to put it back in and see 15 come out.
Given f(x) = x², find f(a − 3).
f(a − 3) = (a − 3)²
the whole bracket is squared
= a² − 6a + 9
(a − 3)(a − 3), all four products
at a = 5: f(2) = 4, and 25 − 30 + 9 = 4 ✓
check with a number
Writing f(a − 3) = a² − 9 squares the a and the 3 separately and loses the middle term −6a. At a = 5 it gives 16, but f(2) is 4 — the number check exposes it at once.
A fraction input: f(x) = 6x − 1, find f(12).
f(12) = 6(12) − 1
brackets as always
= 3 − 1 = 2
6 halves are 3
A fraction goes in exactly like a whole number; multiply first, then subtract.
Backwards with a square: f(x) = x² − 1 and f(k) = 15.
k² − 1 = 15
set the rule equal to 15
k² = 16
add 1
k = 4 or k = −4
both square to 16
Two different inputs give the same output, which a function allows. Give both values of k.
No x in the rule: f(x) = 7, find f(−5).
f(−5) = 7
there is no x to replace
A constant function sends every input to the same output. Its graph is a horizontal line, which passes the vertical line test.
Fares, tax brackets and delivery charges nearly all work this way.
₱25 for the first 2 km
flat, so f(x) = 25 for 0 < x ≤ 2
₱8 for each km after 2
f(x) = 25 + 8(x − 2) for x > 2
f(6) = 25 + 8(4) = 57
the opening problem
The (x − 2) is doing the real work: it charges for the kilometres beyond the first two, not all of them. Every piecewise fare has a subtraction like this somewhere, and leaving it out is what produces ₱73.
Periodical exams ask for f of a number, a negative or an expression; entrance tests like to run it backwards (given f(k), find k) or dress it as a fare or a bill. The fastest route is to write the substitution in full with brackets, then simplify. For a 10-second check, put a small number such as 1 or 5 into both the question and your answer: they must agree.
Evaluate
f(x) = 2x² − 3x + 1, find f(−2)
f(−2) = 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 15. A drops the brackets and reads −2² as −4, giving 2(−4) − 3(−2) + 1 = −1 — and then loses the sign on the middle term too. C substitutes +2 instead of −2. The two sign traps are the squared term and the −3x, and both flip the same way.
Hint: write f(−2) = 2(−2)² − 3(−2) + 1 in full before touching a calculator. Both minus signs are inside brackets, and both make the term positive.
