IGCSE Maths Functions & differentiation notes

Extended. Writing f(x) = 2x + 3 names the machine *f* and its rule. Then f(5) = 2(5) + 3 = 13 — just substitution. The domain is the set of allowed inputs: for f(x) = 1⁄(x − 2), every x except 2 works, so the domain is x ≠ 2. The range is the set of outputs: x² + 2 is a square plus two, so it never falls below 2 — the range is y ⩾ 2.

Quadratic functions pair each input with one output but are not one-one — f(3) = f(−3) for x² — which is why inverses exist cleanly only for one-one functions like 2x + 3. Domain restrictions (x ⩾ 0) can make a quadratic invertible, but at this level inverses are asked for linear and simple rational machines.

Extended. The composite gf(x) means g(f(x)) — the letter next to x acts first. With f(x) = x + 1 and g(x) = x², gf(4) = g(5) = 25, but fg(4) = f(16) = 17. Different orders, different machines. To solve gf(x) = 51 with f(x) = x² + 1 and g(x) = 3x: 3(x² + 1) = 51 → x² = 16 → x = ±4.

To find an inverse, write y = f(x), rearrange to make x the subject, then swap the letters. For f(x) = 3x − 1: y = 3x − 1 → x = (y + 1)⁄3, so f⁻¹(x) = (x + 1)⁄3. The check is free: f⁻¹(f(x)) should return x. Graphically, f and f⁻¹ mirror each other in the line y = x.

Extended. Differentiation gives a formula for the gradient of a curve. The power rule: for axⁿ, the derivative is anxⁿ⁻¹ — multiply by the power, then reduce it. So 5x³ differentiates to 15x², 4x² − 3x + 7 differentiates to 8x − 3 (constants vanish), and sums differentiate term by term. Fractions convert first: 6⁄x = 6x⁻¹, which differentiates to −6x⁻² = −6⁄x².

The derivative at a point is the gradient there — the same number a tangent’s slope would have. For y = x², dy⁄dx = 2x, so at x = 3 the gradient is 6: the curve climbs six units for every one across. Negative gradients mean falling; dy⁄dx = 0 marks a flat instant — the stationary points of the next section.

Extended. Stationary points solve dy⁄dx = 0. For y = x² − 8x + 1: dy⁄dx = 2x − 8 = 0 gives x = 4, and y(4) = −15, so the point is (4, −15). To classify it, use the second derivative: differentiate again to get d²y⁄dx² = 2, which is positive — a minimum. Negative second derivative means a maximum. Alternatively substitute x = 3 and x = 5 into dy⁄dx: the gradient passes −2, 0, +2, so the curve falls, flattens, rises — a valley.

A tangent at x = a has gradient m = dy⁄dx at a and passes through the curve point: for y = x² at (2, 4), m = 4 and the tangent is y − 4 = 4(x − 2), i.e. y = 4x − 4. One step beyond the syllabus, the normal is the perpendicular line: its gradient is −1⁄m (here −1⁄4), because perpendicular gradients multiply to −1. The 0580 papers ask for tangents, so treat the normal as enrichment.