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Maximum and Minimum in a Closed Interval

The absolute maximum and absolute minimum of a continuous function on a closed interval are the greatest and least function values attained on that interval.

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Student-friendly explanation

For a continuous function on a closed interval [a, b], the absolute maximum and absolute minimum are decided by actual values, not by derivative tests alone. First find all candidate x-values: the two endpoints and every critical point lying strictly inside the interval. Then evaluate f(x) at each candidate and compare the numbers. This method is different from local extrema because an endpoint can be the absolute maximum or minimum even though f'(x) may not be zero there. Critical points outside [a, b] are irrelevant, and a local maximum inside the interval may still be smaller than an endpoint value.

How to write this in exams

  1. 1

    Start with the exact idea

    The absolute maximum and absolute minimum of a continuous function on a closed interval are the greatest and least function values attained on that interval.

  2. 2

    Then show how to use it

    Confirm the interval is closed; differentiate; find critical points inside the interval only; evaluate f at endpoints and valid critical points; compare values; write both value and location.

  3. 3

    Add one concrete example

    For f(x) = x^2 - 4x + 1 on [0, 3], f'(x) = 2x - 4, so x = 2. Values: f(0) = 1, f(2) = -3, f(3) = -2. Hence absolute maximum is 1 at x = 0 and absolute minimum is -3 at x = 2.

  4. 4

    Avoid this incomplete answer

    Including a critical point outside the given interval in the comparison table.

Definition

The absolute maximum and absolute minimum of a continuous function on a closed interval are the greatest and least function values attained on that interval.

Example

For f(x) = x^2 - 4x + 1 on [0, 3], f'(x) = 2x - 4, so x = 2. Values: f(0) = 1, f(2) = -3, f(3) = -2. Hence absolute maximum is 1 at x = 0 and absolute minimum is -3 at x = 2.

Rule to remember

Closed interval method: for continuous f on [a, b], evaluate f at a, b, and all critical points c in (a, b) where f'(c) = 0 or f'(c) is undefined. Compare these values to find absolute maximum and minimum.

Memory hook

Closed interval means closed checklist: left endpoint, inside critical points, right endpoint.

Examples and method

Worked example

Find absolute maximum and minimum of f(x) = x^3 - 3x + 2 on [-2, 2]. f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), so critical points in the interval are x = -1 and x = 1. Evaluate: f(-2) = -8 + 6 + 2 = 0, f(-1) = -1 + 3 + 2 = 4, f(1) = 1 - 3 + 2 = 0, f(2) = 8 - 6 + 2 = 4. Absolute maximum is 4 at x = -1 and x = 2; absolute minimum is 0 at x = -2 and x = 1.

Method to apply

Confirm the interval is closed; differentiate; find critical points inside the interval only; evaluate f at endpoints and valid critical points; compare values; write both value and location.

Diagram support

A diagram is not compulsory; a small table of x-values and f(x)-values is usually more exam useful than a graph.

How CBSE asks it

Commonly asked as a direct problem on a given interval, often requiring students to show all candidate values before comparison.

Avoid common mistakes

Common confusion

Students often stop after solving f'(x) = 0 and declare that value as the maximum or minimum. In a closed interval, this can be wrong because the largest or smallest function value may occur at an endpoint, and only a comparison table can confirm the absolute extrema.

Common wrong answer

Including a critical point outside the given interval in the comparison table.

Exam tip

Create a value table with endpoints and valid critical points; comparison of f-values earns the final decision.

Quick check

For f(x) = x^2 on [-2, 1], where is the absolute maximum?

Check endpoints and critical point x = 0. f(-2) = 4, f(0) = 0, f(1) = 1. Absolute maximum is 4 at x = -2.

Answer writing and exam use

1-mark answer

The absolute maximum and absolute minimum of a continuous function on a closed interval are the greatest and least function values attained on that interval.

2-mark answer

The absolute maximum and absolute minimum of a continuous function on a closed interval are the greatest and least function values attained on that interval. Closed interval method: for continuous f on [a, b], evaluate f at a, b, and all critical points c in (a, b) where f'(c) = 0 or f'(c) is undefined. Compare these values to find absolute maximum and minimum. For f(x) = x^2 - 4x + 1 on [0, 3], f'(x) = 2x - 4, so x = 2. Values: f(0) = 1, f(2) = -3, f(3) = -2. Hence absolute maximum is 1 at x = 0 and absolute minimum is -3 at x = 2.

3-mark answer

For a continuous function on a closed interval [a, b], the absolute maximum and absolute minimum are decided by actual values, not by derivative tests alone. First find all candidate x-values: the two endpoints and every critical point lying strictly inside the interval. Then evaluate f(x) at each candidate and compare the numbers. This method is different from local extrema because an endpoint can be the absolute maximum or minimum even though f'(x) may not be zero there. Critical points outside [a, b] are irrelevant, and a local maximum inside the interval may still be smaller than an endpoint value. Closed interval method: for continuous f on [a, b], evaluate f at a, b, and all critical points c in (a, b) where f'(c) = 0 or f'(c) is undefined. Compare these values to find absolute maximum and minimum. Find absolute maximum and minimum of f(x) = x^3 - 3x + 2 on [-2, 2]. f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), so critical points in the interval are x = -1 and x = 1. Evaluate: f(-2) = -8 + 6 + 2 = 0, f(-1) = -1 + 3 + 2 = 4, f(1) = 1 - 3 + 2 = 0, f(2) = 8 - 6 + 2 = 4. Absolute maximum is 4 at x = -1 and x = 2; absolute minimum is 0 at x = -2 and x = 1. Commonly asked as a direct problem on a given interval, often requiring students to show all candidate values before comparison. Including a critical point outside the given interval in the comparison table.
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