ZIMSEC A Level · Pure Mathematics · Form 5
Functions
A function maps each element of the domain to exactly one element of the range.
- Composite function: fg(x) = f(g(x)) – apply g first.
- Inverse function f⁻¹ exists only if f is one-to-one. Swap x and y and make y the subject. The graph of f⁻¹ is the reflection of f in y = x; the domain of f⁻¹ is the range of f.
Worked example: f(x) = 2x + 3 and g(x) = x², x ∈ ℝ. Find fg(x), gf(x) and f⁻¹(x).
fg(x) = 2x² + 3
gf(x) = (2x + 3)²
y = 2x + 3 → x = (y − 3)/2, so f⁻¹(x) = (x − 3)/2
Transformations of graphs
- y = f(x) + a: translation by a units up. y = f(x + a): translation by a units LEFT.
- y = af(x): stretch parallel to the y-axis, factor a. y = f(ax): stretch parallel to the x-axis, factor 1/a.
- y = −f(x): reflection in the x-axis. y = f(−x): reflection in the y-axis.
Modulus function
|x| is the distance from 0. To sketch y = |f(x)|, reflect parts below the x-axis upwards. Solve |x − 2| < 3 as −3 < x − 2 < 3, giving −1 < x < 5.
Exam tip: State the domain whenever you find an inverse – it is often worth a mark.
Check yourself
- f(x) = 3x − 1 and g(x) = 1/x (x ≠ 0). Find gf(x).
- Describe the transformation from y = x² to y = (x − 4)² + 1.
- Solve |2x + 1| = 5.
Show answers
- 1/(3x − 1), x ≠ 1/3
- Translation 4 units right and 1 unit up (vector (4, 1)).
- 2x + 1 = 5 or 2x + 1 = −5, so x = 2 or x = −3
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