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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
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The graph shows y = f′(x), the straight line through (0, 6) and (2, 0). Write the values of x for which f is:
(a) increasing;
- f′ = 0 at x = 2
- f′ is above the axis for \(x<2\)
(b) decreasing.
- f′ is below the axis for \(x>2\)
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The graph shows y = f′(x), a parabola meeting the x-axis at x = −1 and x = 3.
(a) Write the values of x for which f is increasing.
- f is increasing where f′ is above the axis
- \(-1<x<3\)
(b) Write the values of x where f has a stationary point.
- Stationary points are where f′ = 0: x = −1 and 3
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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.
(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: \(2<x<5\)
(b) Write the x-value of the stationary point of f.
- Stationary points need f′ = 0
- x = 5
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The graph shows y = f′(x), a cubic that meets the x-axis at x = −1, 2 and 4.
(a) Write the values of x for which f is increasing.
- f is increasing where f′ is above the axis
- \(-1<x<2\) or \(x>4\)
(b) Write the values of x where f has a stationary point.
- Stationary points where f′ = 0: x = −1, x = 2, x = 4
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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.
(a) Find the x-coordinate of the stationary point of f.
- f′ = 0 at x = 5
(b) Is it a maximum or a minimum?
- f′ is positive then negative either side of x = 5
- f rises then falls: a maximum
(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
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The graph shows y = f′(x) for x > 0. It is a decreasing curve that gets closer to the x-axis.
(a) Find the gradient of f at x = 3.
- Read the height of the f′ graph at x = 3
- f′(3) = 4
(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
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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.
(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 \(5<t\le 6\)
(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
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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.
- 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
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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.
- 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