Maximum and Minimum Values of a Function in a Closed Interval

When a continuous function is defined on a closed interval [a, b], it will always have a highest point and a lowest point within that specific range. Finding these points is crucial because a function might not have a maximum or minimum on an open interval, but closing the interval guarantees these absolute extreme values exist.

The absolute greatest or least value a function achieves over an entire closed interval [a, b]. They are also known as the global maximum or global minimum.

A continuous function on a closed interval [a, b] will attain its absolute maximum and absolute minimum value at least once in that interval.

If a differentiable function has an absolute max or min at an interior point c of the interval, then its derivative at that point is zero (f'(c) = 0).

• Find critical points: Find all points x where f'(x) = 0 or where f is not differentiable.

• Identify endpoints: Note the endpoints a and b of the closed interval.

• Calculate values: Evaluate the original function f(x) at all the critical points and endpoints.

• Compare: The highest value from Step 3 is the absolute maximum, and the lowest is the absolute minimum.

Find the absolute maximum and minimum values of a function \[f\] given by

Solution: We have

or \[f'(x) = 16x^{\frac{1}{3}} - \frac{2}{x^{\frac{2}{3}}} = \frac{2(8x - 1)}{x^{\frac{2}{3}}}\]

Thus, \[f'(x) = 0\] gives \[x = \frac{1}{8}\]. Further note that \[f'(x)\] is not defined at \[x = 0\]. So the critical points are \[x = 0\] and \[x = \frac{1}{8}\]. Now evaluating the value of \[f\] at critical points \[x = 0, \frac{1}{8}\] and at end points of the interval \[x = -1\] and \[x = 1\], we have

\[f(-1) = 12(-1)^{\frac{4}{3}} - 6(-1)^{\frac{1}{3}} = 18\]

\[f(0) = 12(0) - 6(0) = 0\]

Hence, we conclude that absolute maximum value of \[f\] is 18 that occurs at \[x = -1\] and absolute minimum value of \[f\] is \[\frac{-9}{4}\] that occurs at \[x = \frac{1}{8}\].

• Continuity on a closed interval guarantees existence of absolute extrema.

• Differentiability at an interior extremum implies \(f'(c)=0\).

• Endpoints must always be checked in closed interval problems.

• Local extrema and absolute extrema are not always the same.

• A critical point occurs when \(f'(x)=0\) or \(f'(x)\) is undefined.

• Optimisation problems in calculus are applications of maxima and minima.