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Course: Pure Mathematics — Second Secondary
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Curriculum

Pure Mathematics — Second Secondary

Text lesson

Monotonicity of Functions

Pure Mathematics — Second Secondary

Monotonicity of Functions

Worked examples

Worked examples

Example 1: a linear increasing function

  • Show that f(x) = 2x + 1 is increasing on the set of real numbers.
  • Solution Take a = 1 and b = 2. Then f(1) = 3 and f(2) = 5, and the second value is larger. For any a < b, f(b) − f(a) = 2(b − a) > 0 because b − a > 0. So f is increasing on all real numbers.

Example 2: a linear decreasing function

  • Study the monotonicity of g(x) = 4 − 3x.
  • Solution g(a) − g(b) = (4 − 3a) − (4 − 3b) = 3(b − a) > 0 for every a < b. So g(a) > g(b), and g is decreasing on all real numbers.

Example 3: a quadratic function

  • Study the monotonicity of f(x) = x² − 4x + 1.
  • Solution Here a = 1 > 0, so the parabola opens upward. The vertex is at x0 = −(−4) / (2 × 1) = 2. Therefore f is decreasing on ]−∞, 2] and increasing on [2, ∞[.
Check with values→f(0) = 1, f(2) = −3 and f(4) = 1. The values fall until x = 2 and then rise, which confirms the result.

Example 4: an absolute value function

  • Study the monotonicity of h(x) = |x − 1|.
  • Solution When x < 1, h(x) = 1 − x, which is decreasing. When x ≥ 1, h(x) = x − 1, which is increasing. So h is decreasing on ]−∞, 1] and increasing on [1, ∞[.

Example 5: a function defined by two rules

  • Let k(x) = x² when x ≤ 0, and k(x) = 2x when x > 0. Study its monotonicity.
  • Solution On ]−∞, 0] the rule x² is decreasing, because as x increases toward zero, x² decreases. On ]0, ∞[ the rule 2x is increasing. Since k(0) = 0, the function is decreasing on ]−∞, 0] and increasing on [0, ∞[.

Example 6: finding a constant that gives a required monotonicity

Comparison aspectMain ideaKeywords
•
  • Let f(x) = x² + kx. Find the values of k for
  • f is increasing on [3
Findvaluesincreasing
SolutionThe parabola opens upward and its vertex is at x0 = −k / 2. For f to be increasing on [3, ∞[, the vertex must lie at 3 or before it: −k / 2 ≤ 3, so −k ≤ 6 and k ≥ −6.parabolaopensupwardvertex
Check
  • For k = −6
  • the vertex is at 3 and f(x) = x² − 6x is increasing on [3
  • ∞[. For k = −8
  • the vertex is at 4
  • so f decreases between 3 and 4 and the condition fails
vertexincreasingdecreasesbetween
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