Calculus Practice Questions for Matric

Practice questions with solutions

Differentiate: f(x) = x⁴ − 2x² + 5

1. f'(x) = 4x³ − 4x

Answer: 4x³ − 4x

Find the equation of the tangent to y = x² at x = 3

1. dy/dx = 2x → slope = 6
2. Point (3, 9)
3. y − 9 = 6(x − 3)

Answer: y = 6x − 9

Find the stationary points of f(x) = 2x³ − 6x

1. f′(x) = 6x² − 6 = 0
2. x² = 1, so x = 1 or x = −1
3. f(1) = −4 and f(−1) = 4
4. f″(x) = 12x: positive at x = 1, negative at x = −1

Answer: (1; −4) is a local minimum and (−1; 4) a local maximum

Find dy/dx if y = 3√x + 2/x

1. Rewrite with exponents: y = 3x^(1/2) + 2x⁻¹
2. Apply the power rule to each term

Answer: dy/dx = 3/(2√x) − 2/x²

Find f′(x) from first principles if f(x) = x² + 1

1. f(x + h) − f(x) = (x + h)² + 1 − x² − 1 = 2xh + h²
2. Divide by h: 2x + h
3. Let h → 0, keeping 'lim' on every line until h is gone

Answer: f′(x) = 2x

Find the x-coordinates of the turning points of f(x) = x³ − 3x² − 9x + 2.

1. f′(x) = 3x² − 6x − 9 = 0
2. x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0

Answer: x = 3 or x = −1

For which values of x is f(x) = x³ − 3x² − 9x + 2 decreasing?

1. f is decreasing where f′(x) < 0
2. 3(x − 3)(x + 1) < 0

Answer: −1 < x < 3

A ball's height is h(t) = 20t − 5t² metres after t seconds. Find its maximum height.

1. h′(t) = 20 − 10t = 0, so t = 2
2. h(2) = 40 − 20 = 20

Answer: 20 m, after 2 seconds

Tips

• Always show the derivative step.
• Sketch cubic graphs to visualise turning points.