Increasing & Decreasing Functions; First Derivative Test

Definitions

Decreasing Intervals:    x₁ < x₂ implies f(x₁) > f(x₂)

This function is increasing over -oo < x < -2 and
                                                               3 < x < oo and is decreasing over -2 < x < 3

Test for Increasing or Decreasing Intervals

• If f '(x) > 0, then f is increasing
• If f '(x) < 0, then f is decreasing
• If f '(x) = 0, then f is constant

Assume f '(x) > 0 for all x in (a,b).

     I.  Let f '(x) > 0 for all x in (a,b).

          Let x₁ < x₂
            The Mean Value Theorem says
             f '(c) = f(x₂) - f(x₁)
                             x₂ - x₁

            Since f '(x) > 0 and x₂ - x₁ > 0, then f(x₂) - f(x₁) > 0
          Therefore f(x₂) > f(x₁) and f(x) is increasing

    II.  Let f '(x) < 0 for all x in (a,b).

          The proof is similar to the proof in I.

First Derivative Test

1. If f '(x) changes from negative to positive at c, then f(c) is a relative minimum of f(x).
2. If f '(x) changes from positive to negative at c, then f(c) is a relative maximum of f(x).
3. If f '(x) does not change signs at c, then f(c) is neither a relative minimum nor a relative maximum.

     I.  Assume f '(x) changes from negative to positive at c.
         Then f '(x) < 0 for all x in (a,c) and f '(x) > 0 for all x in (c,b).
         Thus f(x) decreases on (a,c) and increases on (c,b).
         Therefore, f(c) is a minimum.

   II.  Assume f '(x) changes from positive to negative at c.
         The proof is similar to the proof in I.

Definition

Problems