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Maximum and Minimum Values of a Function in a Closed Interval

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Estimated time: 10 minutes
CBSE: Class 12

Introduction

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.

CBSE: Class 12

Definition: Absolute Maximum/Minimum

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.

CBSE: Class 12

Theorem: Existence

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

CBSE: Class 12

Theorem: Interior Points

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).

CBSE: Class 12

Working Rule

  • 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.

CBSE: Class 12

Example 1

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

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

Solution:

Step 1: Differentiate the function

We have

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

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

Step 2: Find critical points

2(8x - 1) = 0
8x - 1 = 0
\[x = \frac{1}{8}\]

Also, f'(x) is not defined at x = 0, so x = 0 is also a critical point.

Critical points:

\[x = 0, \quad x = \frac{1}{8}\]
 
Critical points and Endpoints
x = −1, 0, \[\frac{1}{8}\], 1

Step 3: Find function values

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

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

\[f\left(\frac{1}{8}\right) = 12\left(\frac{1}{8}\right)^{\frac{4}{3}} - 6\left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{-9}{4}\]
\[f(1) = 12(1)^{\frac{4}{3}} - 6(1)^{\frac{1}{3}} = 6\]

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 \[x = \frac{1}{8}\].

CBSE: Class 12

Key Points: Maximum and Minimum Values of a Function in a Closed Interval

  • 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.

Video Tutorials

Shaalaa.com | Maxima Minima Part 6 (Maximum and Minimum Value in Closed Interval)

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Maxima Minima Part 6 (Maximum and Minimum Value in Closed Interval) [00:34:03]
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