From f to f′ Problem set A Worked solutions

The graph shows the straight line y = f(x) through (0, 4) and (2, 0).

1. (a) Write down f′(x).

   • Gradient = (0 − 4) ÷ (2 − 0) = −2
   • f′(x) = −2

2. (b) Does the graph of f′ cross the x-axis? Give a reason.

   • The graph of f′ is a horizontal line at y = −2
   • It is never zero, so it does not cross the x-axis

The graph shows y = f(x), a parabola with its vertex at (1, 4). The graphs P, Q and R are shown with it. One of them is the graph of f′.

1. (a) Which of P, Q, R is the graph of f′?

   • f′ is a straight line, and f rises then falls, so f′ is positive then negative: sloping down
   • It is Q

2. (b) State the x-value where the graph of f′ meets the x-axis.

   • f′ = 0 at the vertex, x = 1

The graph shows y = f(x), a cubic with its turning points marked. The graphs P, Q and R are shown with it. One of them is the graph of f′.

1. (a) Which of P, Q, R is the graph of f′?

   • The turning points are at x = −1 and 1, so f′ meets the x-axis there
   • f rises, falls, rises, so f′ is positive, negative, positive: a U shape
   • It is R

2. (b) Write down the values of x where f′(x) = 0.

   • f′ = 0 at the turning points of f
   • x = −1 and x = 1

The graph shows y = f(x), a quartic with its three turning points marked.

1. (a) Write down the values of x where the graph of f′ meets the x-axis.

   • The graph of f′ meets the x-axis at the turning points
   • x = −1, x = 1, x = 3

2. (b) Write down the values of x for which f′(x) > 0.

   • f′ is positive where f rises
   • f′ > 0 for \(-1<x<1\) or \(x>3\)

The graph shows y = f(x) for 0 ≤ x ≤ 360, where x is in degrees. Its maxima and minima are marked.

1. (a) Write down the values of x where f′(x) = 0.

   • f′ = 0 at the turning points
   • x = 90 and x = 270

2. (b) Write down the values of x for which f′(x) > 0.

   • f falls to 90, rises to 270, then falls: f′ is positive for 90 < x < 270

The table shows f(x) = xᵃ and its gradient function for a = 1 to 4. Use the pattern to write f′(x) for a = 5.

• The old power comes to the front, and the power drops by 1
• x⁵ → 5x⁴

A ball is thrown upwards. The graph shows its height h metres after t seconds, for 0 ≤ t ≤ 4. The graphs P, Q and R are shown with it. One of them shows the rate of change of h (the vertical speed).

1. (a) Which of the three graphs shows the rate of change?

   • The rate of change is the gradient, which is f′
   • The graph rises then falls, so the gradient is positive then negative: R

2. (b) State the value of t where the rate of change is zero and say what it means.

   • The gradient is zero at the turning point, t = 2
   • The ball is at its highest point: it is neither rising nor falling for an instant

The graph of y = f(x) is a cubic with a local minimum at (−2, −20) and a local maximum at (1, 7). It falls, rises, then falls. Describe the graph of f′: its shape, where it meets the x-axis and where it is above the axis.

• Turning points of f are x-intercepts of f′: x = −2 and x = 1
• f rises between them, so f′ > 0 for −2 < x < 1
• f falls outside, so f′ < 0 there
• Negative, positive, negative is an upside-down U: a downward parabola

The graph of y = f(x) for 0 ≤ x ≤ 360 is a wave with maxima at (0, 4) and (360, 4) and a minimum at (180, −2). Describe the graph of f′: where it is zero, where it is negative and where it has its smallest and largest values.

• f′ = 0 at the turning points: x = 0, 180, 360
• f falls from 0 to 180: f′ < 0; f rises from 180 to 360: f′ > 0
• f is steepest (falling) at x = 90 and steepest (rising) at x = 270
• So f′ has a minimum at x = 90 and a maximum at x = 270