medium difficulty11 min

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.

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

Practice This Concept

Understand it

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.

Example

For f(x) = x² - 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.

Formula and rule

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.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Find absolute maximum and minimum of f(x) = x³ - 3x + 2 on [-2, 2]. f'(x) = 3x² - 3 = 3(x - 1)(x + 1), so critical points in the interval are x = -1 and x = 1.

1Evaluate: f(-2) = -8 + 6 + 2 = 0, f(-1) = -1 + 3 + 2 = 4, f(1) = 1 - 3 + 2 = 0, f(2) = 8 - 6 + 2 = 4.
2Absolute maximum is 4 at x = -1 and x = 2; absolute minimum is 0 at x = -2 and x = 1.

Method you can reuse

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.

Visual tip

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

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

For f(x) = x² on [-2, 1], where is the absolute maximum?

Think first, then show answer

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

Common wrong answer

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

Don't mix this up

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.

Exam tip

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

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

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

1
Enough for 1 mark
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
Add for 2 marks
Example: For f(x) = x² - 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
Add for 3 marks
First find all candidate x-values: the two endpoints and every critical point lying strictly inside the interval.

MCQ Quiz

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