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IB Mathematics: Applications and Interpretation HL · 12.9: Differentiation · 12.9.4

Sketching f and f' from graphs Problem set A Worked solutions

Skills

  1. The graph shows \(y=f(x)\), a parabola with roots at \(x=-1\) and \(x=5\). Sketch \(y=f'(x)\).

    −2246−6−4−2246810xyy = f(x)
    • Turning point halfway between the roots: x = 2, so f'(2) = 0
    • f rises then falls, so f' is positive then negative
    • f'(x) = −2x + 4: a line through (2, 0) and (0, 4)
  2. The graph shows \(y=f(x)\), a parabola that touches the x-axis at \(x=3\). Sketch \(y=f'(x)\).

    −11234567−224681012xyy = f(x)
    • f touches the axis at x = 3: a turning point, so f'(3) = 0
    • f falls for x < 3 and rises for x > 3, so f' is negative then positive
    • f'(x) = 2x − 6: a line through (3, 0) and (0, −6)
  3. The graph shows \(y=f(x)\), a cubic with roots at \(x=-2\), \(0\) and \(3\). Sketch \(y=f'(x)\), giving the x-values where it crosses the x-axis.

    −3−2−11234−10−5510xyy = f(x)
    • f has a maximum between −2 and 0 and a minimum between 0 and 3
    • From the calculator: turning points at x = −1.12 and x = 1.79
    • f' is positive, negative, positive: an upward parabola with those roots
    • f'(0) = −6, the gradient of f at the origin
  4. The graph shows \(y=f(x)\), a cubic that touches the x-axis at \(x=-1\) and crosses it at \(x=2\). Sketch \(y=f'(x)\).

    −3−2−11234−8−6−4−22468xyy = f(x)
    • The touching point x = −1 is a turning point, so f'(−1) = 0
    • The other turning point is the minimum between −1 and 2, at x = 1
    • f' is positive, negative, positive: upward parabola with roots −1 and 1
  5. The graph shows \(y=f(x)\), an increasing curve with the horizontal asymptote \(y=2\). It crosses the x-axis at \(x=0.405\). Sketch \(y=f'(x)\).

    −2−11234−8−6−4−224xyy = f(x)
    • f is increasing for all x, so f'(x) > 0: the graph is above the x-axis
    • The asymptote of f means f'(x) → 0 for large x
    • f gets steeper to the left, so f' is larger there: f'(0) = 3
  6. Sketch a possible graph of a function f that satisfies each condition.

    1. (a) \(f(x)>0\) for all x, and \(f'(x)>0\) for all x.

      • f > 0 means the curve is above the x-axis
      • f' > 0 means it rises everywhere: e^x does both
    2. (b) f is always increasing, and \(f'\) is always decreasing.

      • f always increasing: f' > 0
      • f' always decreasing: the gradient shrinks, so the curve bends downwards but keeps rising

In context

  1. The graph shows \(g'(t)\), the rate of change of the height of a plant in cm per day, for \(0\le t\le 6\). It is a parabola with roots at \(t=0\) and \(t=6\). Sketch two possible graphs of the height \(g(t)\).

    −11234567−224681012xyy = g'(t)
    • g'(t) ≥ 0, so g never decreases
    • g' = 0 at t = 0 and t = 6: the graph of g is flat at both ends
    • g' is greatest at t = 3: g is steepest there
    • Both curves have the same shape, one shifted up: g(t) = −t³/3 + 3t² + c, for example c = 0 and c = 2
  2. The graph shows \(y=f(x)=3\sin x\) for \(0\le x\le 2\pi\).

    123456−4−3−2−11234xyy = f(x)
    1. (a) Sketch \(y=f'(x)\).

      • f has a maximum at π/2 and a minimum at 3π/2
      • So f' = 0 at those points, with f'(0) = 3 the steepest rise
      • f'(x) = 3 cos x
    2. (b) Taking this same graph as \(y=f'(x)\), sketch a possible \(y=f(x)\).

      • f' is positive on (0, π), so f rises to a maximum at x = π
      • f' is negative on (π, 2π), so f falls again
      • f(x) = −3cos x + c, a cosine wave turned upside down
    3. (c) Explain why your answer to part (b) is not the only one.

      • f(x) + c has the same gradient as f(x) at every x
      • So the graph can be moved up or down
      • One extra fact, such as a point on the curve, fixes c

Stretch

  1. Let \(f(x)=x^2(x^2-4)\).

    1. (a) Use technology to sketch \(y=f(x)\).

      • From the calculator: graph \(y=x^2(x^2-4)\)
      • Roots −2, 0 (touching), 2; minima at x = ±1.41
    2. (b) Find the values of x for which \(f(x)>0\).

      • Read where the curve is above the x-axis
      • x < −2 or x > 2
    3. (c) Find the values of x for which \(f'(x)<0\).

      • Turning points at x = −1.41, 0, 1.41
      • f falls for x < −1.41 and for 0 < x < 1.41
      • f'(x) < 0 on those intervals
  2. The graph shows four curves, A, B, C and D. They form two pairs: in each pair, one curve is the derivative of the other. Write down the pairs and say which curve is the derivative.

    −2−112−3−2−1123xyABCD
    • A is a cubic with turning points at x = −1 and x = 1, so its derivative has roots at ±1 and is a parabola: B
    • C is a parabola with minimum at x = 0, so its derivative is 0 at x = 0 and positive for x > 0: the straight line D
    • The derivative of A is B, and the derivative of C is D

jmaths.xyz · AI HL · 12.9.4 Sketching f and f' from graphs