CAPE Pure Maths Unit 1 study materials
CAPE Pure Maths Unit 1 study resources

CAPE Pure Maths Unit 1: Functions and Graphs

CAPE Pure Maths Unit 1 · Work through your exam, one topic at a time.

All guide sections

Functions and Graphs

This domain tests the function concept itself: domain and range, one-to-one behaviour, inverse and composite functions, transformations of graphs, and sketching modulus, exponential, logarithmic and rational curves. CAPE items often ask you to state a restriction or read a feature from a sketch, so precision in notation matters as much as the algebra.

Domain and range

Start from restrictions: a denominator cannot be zero, an even root needs a non-negative argument, and a logarithm needs a strictly positive argument. For f(x) = a(x − h)² + k the range is [k, ∞) if a > 0 and (−∞, k] if a < 0. For rational functions, the horizontal asymptote bounds the range.

Composite and inverse functions

fg(x) means f(g(x)); the domain of fg is {x ∈ dom g : g(x) ∈ dom f}. An inverse exists only if f is one-to-one on its stated domain. To find it, write y = f(x), swap x and y, and solve for y. The graph of f⁻¹ is the reflection of f in y = x, and dom f⁻¹ = range f, range f⁻¹ = dom f.

Transformations

  • y = f(x) + a: vertical translation by a.
  • y = f(x + a): horizontal translation a to the left.
  • y = af(x): vertical stretch factor a; y = f(ax): horizontal stretch factor 1/a.
  • y = −f(x): reflection in the x-axis; y = f(−x): reflection in the y-axis.
  • y = |f(x)|: the part below the x-axis is reflected above it.

Rational and modulus graphs

Cancel common factors first: a cancelled factor gives a hole, an uncancelled one a vertical asymptote. Compare degrees of numerator and denominator for the horizontal asymptote (equal degrees: ratio of leading coefficients; numerator of lower degree: y = 0).

Worked example

Find f⁻¹ for f(x) = (2x + 1)/(x − 3), x ≠ 3. Let y = (2x + 1)/(x − 3). Then y(x − 3) = 2x + 1, so yx − 3y = 2x + 1 and x(y − 2) = 3y + 1, giving x = (3y + 1)/(y − 2). Hence f⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2. The restriction x ≠ 2 matches the horizontal asymptote y = 2 of f, confirming that the domain of the inverse equals the range of f.

Traps

  • Writing fg as g(f(x)) instead of f(g(x)).
  • Giving an inverse without restricting its domain.
  • Shifting f(x + a) the wrong way.
  • Confusing a hole (cancelled factor) with a vertical asymptote.