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Year 11 Mathematics · Rates of change and the derivative

Name: ______________________

Date: ____________

1. Find the average rate of change of f(x) = x^2 between x = 0 and x = 4.
- ☐ (a) 4
- ☐ (b) 8
- ☐ (c) 16
- ☐ (d) 2
2. What is the derivative of x^5?
- ☐ (a) x^4
- ☐ (b) 5x^5
- ☐ (c) 5x^4
- ☐ (d) 4x^5
3. What is the derivative of the constant function f(x) = 7?
- ☐ (a) 7
- ☐ (b) 0
- ☐ (c) 7x
- ☐ (d) 1
4. A car's distance travelled, in metres, after t seconds is d(t) = t^2. What is its velocity at t = 3 seconds?
- ☐ (a) 9 m/s
- ☐ (b) 3 m/s
- ☐ (c) 12 m/s
- ☐ (d) 6 m/s
5. As the second point on a secant moves closer to the first point (h approaches 0), what does the gradient of the secant approach?
- ☐ (a) Zero
- ☐ (b) The y-intercept
- ☐ (c) The gradient of the tangent
- ☐ (d) Infinity
6. If f(x) = 3x^2 + 4x, find f'(1).
7. Find the average rate of change of f(x) = 2x + 1 between x = 1 and x = 5.
8. Find the gradient of the tangent to y = x^2 at x = −3.

Answer sheet

1. 4 — (f(4) − f(0)) / (4 − 0) = (16 − 0) / 4 = 4.
2. 5x^4 — Power rule: bring down the 5 and reduce the power by 1, giving 5x^4.
3. 0 — The graph of y = 7 is a horizontal line, which has gradient 0 everywhere.
4. 6 m/s — Velocity is the derivative: d'(t) = 2t, so d'(3) = 6 m/s. (9 m is the distance travelled, not the velocity.)
5. The gradient of the tangent — This limiting process is exactly how the gradient of the tangent, the derivative, is defined.
6. 10 — f'(x) = 6x + 4, so f'(1) = 6 + 4 = 10.
7. 2 — f(1) = 3 and f(5) = 11, so (11 − 3) / (5 − 1) = 8 / 4 = 2. For a straight line, this is just its gradient.
8. −6 — dy/dx = 2x, so at x = −3 the gradient is 2 × (−3) = −6.