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IB Mathematics: Applications and Interpretation HL · 12.9: Differentiation · 12.9.3

From f′ to f Problem set A Worked solutions

Skills

  1. The graph shows y = f′(x), the straight line through (0, 6) and (2, 0). Write the values of x for which f is:

    −2−112345−8−6−4−22468xyy = f′(x)
    1. (a) increasing;

      • f′ = 0 at x = 2
      • f′ is above the axis for \(x<2\)
    2. (b) decreasing.

      • f′ is below the axis for \(x>2\)
  2. The graph shows y = f′(x), a parabola meeting the x-axis at x = −1 and x = 3.

    −3−2−112345−6−4−2246xyy = f′(x)
    1. (a) Write the values of x for which f is increasing.

      • f is increasing where f′ is above the axis
      • \(-1<x<3\)
    2. (b) Write the values of x where f has a stationary point.

      • Stationary points are where f′ = 0: x = −1 and 3
  3. The graph shows y = f′(x). It has asymptotes the dashed lines x = 2 and y = −1 and meets the x-axis at x = 5.

    −22468−4−2246xyy = f′(x)
    1. (a) Write the values of x for which f is increasing.

      • f′ = 0 at x = 5
      • For x < 2 the graph is below y = −1, so below the axis
      • Above the axis only between the asymptote and x = 5
      • Increasing: \(2<x<5\)
    2. (b) Write the x-value of the stationary point of f.

      • Stationary points need f′ = 0
      • x = 5
  4. The graph shows y = f′(x), a cubic that meets the x-axis at x = −1, 2 and 4.

    −2−112345−10−5510xyy = f′(x)
    1. (a) Write the values of x for which f is increasing.

      • f is increasing where f′ is above the axis
      • \(-1<x<2\) or \(x>4\)
    2. (b) Write the values of x where f has a stationary point.

      • Stationary points where f′ = 0: x = −1, x = 2, x = 4
  5. The graph shows y = f′(x), the line through (0, 10) and (5, 0). A different function g has g′(x) = 3 for all x.

    −22468−5510xyy = f′(x)
    1. (a) Find the x-coordinate of the stationary point of f.

      • f′ = 0 at x = 5
    2. (b) Is it a maximum or a minimum?

      • f′ is positive then negative either side of x = 5
      • f rises then falls: a maximum
    3. (c) Describe the graph of g.

      • g′ is the constant 3, so the gradient of g never changes
      • g is a straight line, and it is never stationary
  6. The graph shows y = f′(x) for x > 0. It is a decreasing curve that gets closer to the x-axis.

    123456782468xyy = f′(x)
    1. (a) Find the gradient of f at x = 3.

      • Read the height of the f′ graph at x = 3
      • f′(3) = 4
    2. (b) Is f increasing or decreasing? Does its gradient get larger or smaller as x increases?

      • f′ > 0 for all x > 0, so f is increasing
      • f′ decreases towards 0, so the gradient of f gets smaller

In context

  1. The graph shows the rate of change of a population of bacteria, P′(t) thousand per hour, for 0 ≤ t ≤ 6, where P is the population in thousands and t is the time in hours. P′ meets the t-axis at t = 1 and t = 5.

    123456−6−4−2246t (hours)P′P′(t)
    1. (a) For which values of t is the population decreasing?

      • The population is decreasing where P′ is below the axis
      • \(0\le t<1\) or \(5<t\le 6\)
    2. (b) At what time is the population at a local maximum?

      • P′ changes from positive to negative at t = 5
      • The population is largest there: t = 5

Stretch

  1. The graph of y = f′(x) is the line through (0, 8) and (4, 0). Describe the graph of y = f(x): where it has a stationary point, whether that is a maximum or a minimum, and its overall shape.

    −2−1123456−4−2246810xyy = f′(x)
    • f′ = 0 at x = 4
    • f′ > 0 for x < 4: f rises. f′ < 0 for x > 4: f falls
    • Rising then falling: a maximum at x = 4
    • An upside-down U shape, steepest far from x = 4
  2. For x > 0 the graph of y = f′(x) is the decreasing curve through (1, 6), (2, 3) and (6, 1), getting closer to the x-axis. The graph of y = f(x) passes through (1, 2). Describe the graph of f for x > 0.

    123456782468xyy = f′(x)
    • f′ > 0 for all x > 0: f is increasing
    • f′ decreases, so the gradient of f gets smaller: the curve flattens
    • At (1, 2) the gradient is 6; at x = 6 it is 1
    • No stationary point, since f′ is never 0

jmaths.xyz · AI HL · 12.9.3 From the graph of f′ back to f