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

From f′ to f Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. The graph shows y = f′(x), the straight line through (0, −6) and (3, 0). Write the values of x for which f is:

    −2−1123456−8−6−4−2246xyy = f′(x)
    1. (a) increasing;

      • f′ = 0 at x = 3
      • f′ is above the axis for \(x>3\)
    2. (b) decreasing.

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

    −4−3−2−11234−6−4−22468xyy = f′(x)
    1. (a) Write the values of x for which f is increasing.

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

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

    −6−5−4−3−2−112−6−4−2246xyy = f′(x)
    1. (a) Write the values of x for which f is increasing.

      • f′ = 0 at x = −1
      • For x < −3 the graph is below y = −2, so below the axis
      • Above the axis only between the asymptote and x = −1
      • Increasing: \(-3<x<-1\)
    2. (b) Write the x-value of the stationary point of f.

      • Stationary points need f′ = 0
      • x = −1
  4. The graph shows y = f′(x), a cubic that meets the x-axis at x = −3, 0 and 2.

    −4−3−2−1123−10−5510xyy = f′(x)
    1. (a) Write the values of x for which f is increasing.

      • f is increasing where f′ is above the axis
      • \(x<-3\) or \(0<x<2\)
    2. (b) Write the values of x where f has a stationary point.

      • Stationary points where f′ = 0: x = −3, x = 0, x = 2
  5. The graph shows y = f′(x), the line through (0, −6) and (2, 0). A different function g has g′(x) = −2 for all x.

    −2−112345−10−8−6−4−2246xyy = f′(x)
    1. (a) Find the x-coordinate of the stationary point of f.

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

      • f′ is negative then positive either side of x = 2
      • f falls then rises: a minimum
    3. (c) Describe the graph of g.

      • g′ is the constant −2, 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.

    246810−1123456xyy = f′(x)
    1. (a) Find the gradient of f at x = 4.

      • Read the height of the f′ graph at x = 4
      • f′(4) = 2
    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 the water level in a reservoir, L′(t) centimetres per day, for 0 ≤ t ≤ 8, where L is the level in centimetres and t is the time in days. L′ meets the t-axis at t = 2 and t = 7.

    12345678−8−6−4−2246t (days)L′L′(t)
    1. (a) For which values of t is the water level falling?

      • The level is falling where L′ is below the axis
      • \(0\le t<2\) or \(7<t\le 8\)
    2. (b) At what time is the water level at a local maximum?

      • L′ changes from positive to negative at t = 7
      • The level is highest there: t = 7

Stretch

  1. The graph of y = f′(x) is the line through (0, −6) and (3, 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.

    −2−1123456−8−6−4−2246xyy = f′(x)
    • f′ = 0 at x = 3
    • f′ < 0 for x < 3: f falls. f′ > 0 for x > 3: f rises
    • Falling then rising: a minimum at x = 3
    • A U shape
  2. For x > 0 the graph of y = f′(x) is the decreasing curve through (1, 5), (5, 1) and (10, 0.5), getting closer to the x-axis. The graph of y = f(x) passes through (1, 3). Describe the graph of f for x > 0.

    246810−11234567xyy = 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, 3) the gradient is 5; at x = 5 it is 1
    • No stationary point, since f′ is never 0

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