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IB Mathematics: Applications and Interpretation HL · 12.9: Differentiation · 12.9.1

Increasing and decreasing Problem set A Worked solutions

Skills

  1. The graph shows y = f(x), a parabola with its highest point at (2, 4). Write the values of x for which f is increasing.

    −2−1123456−4−224xyy = f(x)(2, 4)
    • Vertex at x = 2: gradient zero
    • The graph rises up to the vertex and falls after it
    • Increasing: \(x<2\)
  2. The graph shows y = f(x), a cubic with turning points at x = −2 and x = 1. Write down the values of x for which f is:

    −4−3−2−1123−101020xyy = f(x)(−2, 10)(1, −3.5)
    1. (a) increasing;

      • Turning points at x = −2 and x = 1 split the axis into three intervals
      • Where the graph rises, f is increasing
      • Increasing: \(x<-2\) or \(x>1\)
    2. (b) decreasing.

      • Where the graph falls, f is decreasing
      • Decreasing: \(-2<x<1\)
  3. The graph shows y = f(x) for 0 ≤ x ≤ 360, where x is in degrees. It has a maximum at (90, 4) and a minimum at (270, −4).

    90180270360−4−224x (degrees)yy = f(x)(90, 4)(270, −4)
    1. (a) Write down the values of x for which f is increasing.

      • Maximum at x = 90, minimum at x = 270
      • The graph rises from 0 to 90 and from 270 to 360
      • Increasing: \(0\le x<90\) or \(270<x\le 360\)
    2. (b) Write down the values of x for which the gradient of the graph is zero.

      • Gradient is zero at the turning points
      • x = 90 and x = 270
  4. The line shown is the tangent to the curve at P(7, 9). It passes through (4, 0) and (6, 6).

    −22468−224681012xyy = f(x)P
    1. (a) Find the gradient of the curve at P.

      • Gradient = rise ÷ run = (6 − 0) ÷ (6 − 4)
      • Gradient at P = 3
    2. (b) Write down the value of x at which the gradient of the curve is zero.

      • The lowest point of the curve is at x = 1
      • The tangent there is horizontal, so the gradient is 0 at x = 1

In context

  1. The graph shows the height h metres of a cyclist above the valley floor after riding d kilometres along a route, for 0 ≤ d ≤ 6. The turning points are (2, 20) and (4, 16).

    123456510152025303540d (km)h (m)h(2, 20)(4, 16)
    1. (a) For which values of d is the cyclist going uphill?

      • Turning points at d = 2 and d = 4
      • Uphill means h is increasing
      • Uphill: \(0\le d<2\) or \(4<d\le 6\)
    2. (b) State the gradient of the graph at d = 4 and say what it tells you about the road.

      • The graph has a turning point at d = 4
      • Gradient 0: the cyclist is neither climbing nor descending at that moment; the road is level

Stretch

  1. A function f is defined for all x. It is increasing for x < −1, decreasing for −1 < x < 3 and increasing for x > 3. Describe the graph of y = f(x): where the gradient is zero, what kind of turning points they are, and the overall shape.

    • Gradient changes sign only at turning points: x = −1 and x = 3
    • At x = −1 the graph changes from rising to falling: local maximum
    • At x = 3 it changes from falling to rising: local minimum
    • Rises, falls, rises: a cubic shape
  2. The graph shows y = f(x), a quartic with a W shape.

    −3−2−11232468xyy = f(x)(−2, 0)(0, 4)(2, 0)
    1. (a) Write down the values of x for which f is increasing.

      • Turning points at −2, 0, 2
      • The graph rises between the first minimum and the maximum, and to the right of the last minimum
      • Increasing: \(-2<x<0\) or \(x>2\)
    2. (b) Write down the values of x at which the gradient is zero.

      • The gradient is zero at every turning point
      • x = −2, x = 0, x = 2

jmaths.xyz · AI HL · 12.9.1 Gradient, increasing and decreasing