Limits; derivative as gradient and rate of change practice questions.

For f(x) = x² − 6x, the gradient of the curve at a certain point equals 4. Determine the x-coordinate of that point.

• x = 5
• x = 1
• x = −5
• x = 2

Answer: x = 5

f′(x) = 2x − 6; setting 2x − 6 = 4 gives 2x = 10, so x = 5.

Given f(x) = 7x − 2, state f′(x).

• 7x
• 7
• −2
• 5

Answer: 7

The derivative of a linear function 7x − 2 is its constant gradient, 7.

The height of a ball is h(t) = 20t − 5t² metres after t seconds. Determine the ball's velocity h′(t) at t = 1 second.

• 15 m s⁻¹
• 10 m s⁻¹
• 20 m s⁻¹
• −10 m s⁻¹

Answer: 10 m s⁻¹

h′(t) = 20 − 10t, so h′(1) = 20 − 10 = 10 m s⁻¹.

Given f(x) = x² + 4x + 3, determine the equation of the gradient function f′(x).

• 2x + 4
• 2x + 3
• x + 4
• 2x² + 4

Answer: 2x + 4

Differentiating: d/dx(x²) = 2x, d/dx(4x) = 4, constant → 0, so f′(x) = 2x + 4.

The tangent to the graph of y = f(x) at the point P(3, 2) also passes through (5, 8). Determine f′(3).

• 3
• 6
• 2
• 8

Answer: 3

f′(3) is the tangent's gradient: (8 − 2)/(5 − 3) = 3. Using only the rise 6, or reading off a y-coordinate, skips the run in the denominator.

A tank is being drained. Its volume V, in litres, t minutes after the tap is opened satisfies V′(20) = −6. Interpret this value in the context of the tank.

• At t = 20 minutes the volume is falling at 6 litres per minute.
• After 20 minutes the tank holds 6 litres less than it did at the start.
• The tank loses on average 6 litres per minute over the first 20 minutes.
• The tank contains −6 litres of water once 20 minutes have passed.

Answer: At t = 20 minutes the volume is falling at 6 litres per minute.

A derivative is an instantaneous rate at one instant — neither a total change nor an average over an interval. The negative sign gives the direction of the change, not a negative volume.