First principles, rules of differentiation, cubic graphs, optimisation and rates of change for NSC Paper 1. AI step-by-step explanations.
Calculus appears in Paper 1 and is usually worth 30–35 marks. CAPS Grade 12 covers first principles, rules of differentiation, cubic graph sketching, optimisation and rates of change. Calculus connects directly to the cubic function: the first derivative gives the stationary points, the second derivative tells you their nature, and the point of inflection lies where f''(x) = 0. In optimisation and rate-of-change questions, the real skill is translating the words into a function you can differentiate.
f'(x) = lim(h→0) [f(x+h) − f(x)] / h = lim(h→0) [3(x+h)² − 2 − (3x² − 2)] / h = lim(h→0) (6xh + 3h²) / h = 6x.
Rewrite as powers: y = 4x³ − 2x⁻¹ + x^(1/2). Then dy/dx = 12x² + 2x⁻² + ½x^(−1/2) = 12x² + 2/x² + 1/(2√x).
f(2) = 8 − 6 = 2, so the point is (2, 2). f'(x) = 3x² − 3, so the gradient is f'(2) = 9. y − 2 = 9(x − 2) gives y = 9x − 16.
f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0, so x = 1 or x = 3. f(1) = 5 and f(3) = 1. f''(x) = 6x − 12: f''(1) < 0 gives a local maximum at (1, 5) and f''(3) > 0 a local minimum at (3, 1). The point of inflection is at x = 2: (2, 3).
Let the length be x, so the width is 20 − x and A = x(20 − x) = 20x − x². A'(x) = 20 − 2x = 0 gives x = 10, and A''(x) = −2 < 0 confirms a maximum. Maximum area = 10 × 10 = 100 cm².
No — Grade 12 Mathematics covers only differential calculus. Integration is not examined.
f'(x) is a function that gives the gradient of the tangent at any x-value. Substituting a particular x-value gives the gradient of the tangent at that point.
Solve f''(x) = 0. For a cubic, the x-coordinate is also exactly halfway between the x-coordinates of the two turning points.
Yes. Write f'(x) = or dy/dx = correctly, and keep 'lim h→0' on every line of a first-principles answer until you actually take the limit.
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