Maxima and minima: critical points and endpoints

Goals and prerequisites

Find and classify extrema of real single-variable functions. Prerequisites: continuity and derivatives. Always state the feasible domain; the same formula can have different extrema on different intervals.

1. Local is different from absolute

An absolute minimum at c satisfies f(c)≤f(x) for every feasible x; a local minimum requires this only in a neighbourhood within the domain. Reverse inequalities for maxima. Strict extrema use strict inequality for other nearby points. In this lesson, derivative-based local tests concern interior points; endpoint extrema are checked separately relative to the domain.

If f is continuous on closed bounded [a,b], the extreme value theorem guarantees absolute extrema. It does not supply their locations. An open interval or an unbounded domain need not attain bounds. For f(x)=x on (0,1), the infimum is 0 and supremum is 1, but neither is attained: there is no absolute minimum or maximum.

2. Candidates and Fermat's condition

At an interior local extremum where f is differentiable, f′(c)=0. This is necessary, not sufficient. For f(x)=x³, f′(0)=0 but f is increasing through 0 and has no extremum there. A critical point in the domain is an interior point where f′=0 or the derivative fails to exist. The second category matters: |x| has a minimum at zero.

For continuous functions on [a,b], list interior critical points and both endpoints, evaluate f at each, and compare values. A point outside the domain is not a candidate. If discontinuities or an unbounded/open domain are present, separately analyze limits and whether bounds are attained; a finite candidate list is not an automatic guarantee.

3. First derivative signs

On intervals where f′>0, f is strictly increasing; where f′<0, it is strictly decreasing. At a continuous interior candidate, a derivative sign change from + to − gives a strict local maximum, and − to + gives a strict local minimum. No sign change gives no extremum under this sign-test setting.

Example: f(x)=x³−3x on [−2,2]. f′=3(x−1)(x+1). Its signs are + for x<−1, − for −1<x<1, and + for x>1. Hence −1 is a local maximum and 1 a local minimum. Evaluate f(−2)=−2, f(−1)=2, f(1)=−2, f(2)=2. Absolute maximum 2 occurs at −1 and 2; absolute minimum −2 occurs at −2 and 1. Absolute extrema need not be unique.

4. Second derivative test and failure cases

If f is twice continuously differentiable near c and f′(c)=0, then f″(c)>0 gives a strict local minimum and f″(c)<0 a strict local maximum. If f″(c)=0, the test is inconclusive. At zero, x⁴ has a strict minimum, −x⁴ a strict maximum, and x³ neither, yet all have first and second derivatives zero there.

For f(x)=x+1/x on x>0, f′=1−1/x² vanishes at 1. It is negative before 1 and positive after 1; f(1)=2 is an absolute minimum, also verified by x+1/x≥2. Since f→+∞ at both domain ends, there is no absolute maximum. Restricting to [2,4] changes the minimum to f(2)=5/2 because the function is increasing there.

5. Modelling a bounded optimization

A rectangle with perimeter 20 has positive side lengths x and 10−x, so 0<x<10. Area A=x(10−x); A′=10−2x, A″=−2. The unique maximum is 25 at x=5. The infimum is 0 but no minimum is attained for nondegenerate rectangles. If degenerate rectangles were allowed, the feasible interval would be [0,10] and the minimum 0 would occur at the endpoints. Domain assumptions alter the answer.

Checklist: translate restrictions first, differentiate within the domain, include nondifferentiable candidates, classify locally, compare globally, and separate an attained extremum from a limiting bound.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

CE01. At x=0, f(x)=x³ has:

A. Strict local maximum / कठोर स्थानीय अधिकतम B. Strict local minimum / कठोर स्थानीय न्यूनतम C. Neither local maximum nor local minimum / कोई स्थानीय चरम नहीं D. Absolute maximum / परम अधिकतम

Answer: C

A: It increases through zero.

B: It also takes negative values just left of zero.

C: The derivative vanishes but does not change sign.

D: It is unbounded above.

CE02. What is the absolute maximum value of x³−3x on [−2,2]?

A. 2 B. −2 C. 0 D. 6

Answer: A

A: Evaluate endpoints and ±1: the largest value is 2.

B: This is the absolute minimum.

C: The function also reaches 2.

D: No candidate has value 6.

CE03. If f′(c)=f″(c)=0, what does the second derivative test conclude?

A. A minimum / न्यूनतम B. A maximum / अधिकतम C. No extremum / कोई चरम नहीं D. Inconclusive / अनिर्णायक

Answer: D

A: x⁴ fits, but −x⁴ does not have a minimum.

B: −x⁴ fits, but x⁴ does not have a maximum.

C: x⁴ provides a counterexample.

D: Higher behaviour or derivative signs must be examined.

CE04. For f(x)=x on (0,1), which statement is true?

A. Minimum 0 and maximum 1 / न्यूनतम 0, अधिकतम 1 B. No absolute minimum or maximum / कोई परम न्यूनतम या अधिकतम नहीं C. Minimum 0 only / केवल न्यूनतम 0 D. Maximum 1 only / केवल अधिकतम 1

Answer: B

A: Neither endpoint belongs to the domain.

B: Every feasible value has both larger and smaller feasible values.

C: Zero is not attained.

D: One is not attained.

CE05. A nondegenerate rectangle has perimeter 20. Its maximum area is:

A. 20 B. 50 C. 25 D. 100

Answer: C

A: At sides 5 and 5 the area is 25.

B: This exceeds the maximum of x(10−x).

C: A′=10−2x vanishes at x=5; A″<0.

D: Sides 10 and 10 would have perimeter 40.

CE06. What is the minimum of x+1/x on [2,4]?

A. 2 B. 17/4 C. 0 D. 5/2

Answer: D

A: The minimizer x=1 for the positive half-line is outside this interval.

B: This is the maximum at x=4.

C: All values here are positive and at least 5/2.

D: The derivative 1−1/x² is positive here; take x=2.

Source and scope

Official Maxima and minima heading; single-variable local and global optimization with explicit domains.

Official GATE 2026 CS syllabus (verified 2026-10-14).