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Application of Derivatives

Absolute maximum and minimum

Find absolute maximum and minimum values on a closed interval.

The absolute (or global) maximum is the highest value of all on the given interval; the absolute minimum is the lowest.

A continuous function on a closed interval [a, b] always has both. It cannot escape: the end points are included.

Working rule (no tests needed):

  • Find all critical points inside the interval: f′ = 0, or f′ does not exist.
  • Add the two end points a and b.
  • Work out f at every one of these points.
  • The largest value is the absolute max; the smallest is the absolute min.

Example: f(x) = x² – 4x on [0, 5]. f′ = 2x – 4 = 0 at x = 2. Values: f(0) = 0, f(2) = –4, f(5) = 5. Absolute max 5 (at x = 5), absolute min –4 (at x = 2).

Note: the maximum here is at an end point, not at a turning point. That is why the end points must always be checked.

Ignore critical points that lie outside the given interval.

Watch for points where f′ does not exist. A function with x^(1/3) has a derivative containing 1/x^(2/3), which breaks at x = 0. So x = 0 is a critical point too, if it is in the interval.

Check every critical point and both ends.

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