One function, two rules, and a condition that says which one you are in.
Given
f(x) = 2x + 1 if x < 2
f(x) = x² − 3 if x ≥ 2
Find f(2).
x = 2 satisfies x ≥ 2, so the second rule owns it: 2² − 3 = 1. D uses the first rule (2(2) + 1 = 5), but that branch requires x < 2 strictly. The equals sign decides who gets the boundary point.
Test the input first. Only then do any arithmetic.
| Input | Which condition holds | Rule to use |
|---|---|---|
| x = 0 | 0 < 2 ✓ | 2x + 1 |
| x = 2 | 2 ≥ 2 ✓ | x² − 3 |
| x = 5 | 5 ≥ 2 ✓ | x² − 3 |
A piecewise function is still one function: every input gets exactly one output. The conditions are written so that they never overlap and never leave a gap.
f(0) = 2(0) + 1
0 is below 2
= 1
the linear piece
f(5) = 5² − 3
5 is at or above 2
= 22
the quadratic piece
Running the input through each rule and adding, or picking whichever looks nicer, are both ways of answering a question nobody asked. One input, one condition, one rule.
Here x ≥ 2 owns the number 2, because that is where the “or equal to” sits.
f(2) = 2² − 3
the ≥ rule wins
= 1
not 5
2(2) + 1 = 5
what the other rule would have given
Swap < and ≤ and the value at the boundary changes completely. It is the one number the question is guaranteed to ask about, and 5 will be sitting there in the choices.
the point is included
from ≤ or ≥
the curve stops just short
from < or >
The jump from 5 down to 1 at x = 2 is real — the graph is broken there. Only one of the two circles is filled, so the function still gives one answer.
Fares, tax brackets, parking, delivery charges — anything priced in bands.
₱40 for the first 3 hours
C(t) = 40 for 0 < t ≤ 3
₱15 for each hour after
C(t) = 40 + 15(t − 3) for t > 3
C(5) = 40 + 15(2)
two hours past the band
= ₱70
not 15 × 5
The extra charge applies to the hours beyond the first three, so the subtraction has to be inside the bracket. Charging 15t would bill the flat-rate hours twice.
A lot charges ₱30 for the first hour or any part of it, then ₱12 for each additional hour or part of an hour. A car stays 4 hours 20 minutes. Then: a driver paid ₱90 — what is the longest stay?
4 h 20 min − 1 h = 3 h 20 min
time beyond the flat band
3 h 20 min → 4 chargeable hours
the 20 minutes start a whole new hour
30 + 12 × 4 = ₱78
not 30 + 12 × 3⅓ = ₱70
Backwards: (90 − 30) ÷ 12 = 5 extra hours
take off the flat band first
1 + 5 = 6 hours at most
one minute more starts a 6th extra hour
Charging the fraction exactly (3⅓ hours) undercharges, and dividing ₱90 by ₱12 forgets the flat first band. Subtract the band, round the rest up.
Three pieces, a negative input. Let f(x) = 1 − x² if x < −1, f(x) = 3 if −1 ≤ x ≤ 2, and f(x) = 2x − 1 if x > 2. Find f(−3), f(2) and f(2.5).
f(−3) = 1 − (−3)² = 1 − 9 = −8
−3 < −1: first piece; square in brackets
f(2) = 3
−1 ≤ 2 ≤ 2: the middle piece owns 2
f(2.5) = 2(2.5) − 1 = 4
2.5 > 2: last piece
What is different: a constant piece ignores x completely, and a middle condition has two ends to test.
Squaring comes before the minus in front: 1 − (−3)² = 1 − 9 = −8. Writing 1 + 9 = 10 lets the minus in front cancel the minus of −3, as if the square were not there.
A function of a function. Let f(x) = x² if x < 3, and f(x) = 10 − 2x if x ≥ 3. Find f(f(2)).
f(2) = 2² = 4
inside first: 2 < 3, the square piece
f(f(2)) = f(4)
the output becomes the new input
f(4) = 10 − 2(4) = 2
4 ≥ 3: now the OTHER piece
What is different: the condition is tested twice, once per input. Using x² again would give 16, the wrong answer.
Let f(x) = x² − 4 if x ≤ 1, and f(x) = 2x + 3 if x > 1. Solve f(x) = 5.
x² − 4 = 5 → x = 3 or x = −3
first piece
keep −3, reject 3
this piece only lives at x ≤ 1
2x + 3 = 5 → x = 1
second piece
reject 1
this piece needs x > 1, and 1 is not
x = −3 only
one solution, not three
Check: f(−3) = 9 − 4 = 5 ✓, while f(1) = 1 − 4 = −3 and f(3) = 2(3) + 3 = 9. The same habit answers “how many solutions” and f(f(x)) questions: work inside out, choosing the rule fresh at every step.
A letter inside a rule works the same way. Let h(x) = mx − 2 if x < 1, and h(x) = x² + m if x ≥ 1, with h(1) = 5. The input 1 lives in the x ≥ 1 piece, so 1 + m = 5 and m = 4. Now use m in the other piece: h(−2) = 4(−2) − 2 = −10.
The boundary value, a fare with “or any part of”, and “how many x give f(x) = c” are the favourites. For each candidate answer, spend two seconds asking “which condition does this x satisfy?” — that alone kills most wrong choices.
Values just below 2 should follow the line; values at 2 and above should follow the parabola.
Given g(x) = x² + 1 if x < −1, and 4 − 3x if x ≥ −1
Find g(−1) + g(−3).
−1 satisfies x ≥ −1, so g(−1) = 4 − 3(−1) = 7. −3 is below −1, so g(−3) = (−3)² + 1 = 10. Sum: 7 + 10 = 17. A gives the boundary to the wrong rule: (−1)² + 1 = 2, and 2 + 10 = 12. C sends −3 through the second rule: 4 − 3(−3) = 13, and 7 + 13 = 20. D drops the sign in 4 − 3(−1), getting 1 instead of 7: 1 + 10 = 11.
