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"Find the absolute maximum value of f on [a, b]" is one of the most reliable questions in Class 12 calculus, and one of the easiest to get wrong by doing too much. Students run the second derivative test, find a local maximum, and report it, while the real answer sits at an endpoint where the derivative is not zero at all. This post sets out the rule, explains why it works, and works three functions where the answer is not where you might first look.
Where is this in the syllabus?
Section 6.4.1 of NCERT's Class 12 chapter on application of derivatives (2026-27 reprint) covers "Maximum and Minimum Values of a Function in a Closed Interval". The JEE (Main) 2026 syllabus lists "Maxima and minima of functions of one variable" in Unit 7, and the CBSE 2026-27 curriculum lists maxima and minima in Class 12. Check your own year's documents.
Why does the interval have to be closed?
NCERT's opening example is f(x) = x + 2 on the open interval (0, 1). It has no maximum and no minimum: whatever value you pick, there is a slightly larger one closer to 1. Close the interval to [0, 1] and the function has a maximum of 3 at x = 1 and a minimum of 2 at x = 0, both at endpoints.
NCERT then states two results without proof. Theorem 5: a continuous function on [a, b] attains an absolute maximum and an absolute minimum at least once. Theorem 6: if f is differentiable and the absolute maximum or minimum is at an interior point c, then f′(c) = 0. Together they say the answer must be at an endpoint or at a critical point, so you only need to check those.
What is the four-step rule?
NCERT's working rule:
Find all critical points in the interval: where f′(x) = 0 or where f is not differentiable.
Take the two endpoints.
Evaluate f at every point from steps 1 and 2.
The largest value is the absolute maximum; the smallest is the absolute minimum.
No second derivative test is needed. You are comparing a short list of numbers.
Worked example 1: a critical point outside the interval
Find the absolute extrema of f(x) = x³ − 12x on [0, 3].
f′(x) = 3x² − 12 = 0 gives x = 2 or x = −2. Only x = 2 lies in [0, 3].
Endpoints: x = 0 and x = 3.
f(0) = 0, f(2) = 8 − 24 = −16, f(3) = 27 − 36 = −9.
Absolute maximum 0 at x = 0; absolute minimum −16 at x = 2.
Including x = −2 would give f(−2) = 16 and a wrong maximum. Always discard critical points outside the interval.
Worked example 2: a trigonometric function
Find the absolute extrema of f(x) = sin x + cos x on [0, π].
f′(x) = cos x − sin x = 0 gives tan x = 1, so x = π/4 in this interval.
f(0) = 1, f(π/4) = √2/2 + √2/2 = √2 ≈ 1.41, f(π) = 0 − 1 = −1.
Absolute maximum √2 at x = π/4; absolute minimum −1 at x = π.
Here the maximum is at the critical point but the minimum is at an endpoint, where f′(π) = −1 is not zero.
Worked example 3: where f′ does not exist
Find the absolute extrema of f(x) = 3x^(2/3) − 2x on [−1, 8].
f′(x) = 2x^(−1/3) − 2. It is zero when x^(1/3) = 1, so x = 1. It does not exist at x = 0, so x = 0 is also a critical point.
f(−1) = 3(1) + 2 = 5. f(0) = 0. f(1) = 3 − 2 = 1. f(8) = 3(4) − 16 = −4.
Absolute maximum 5 at x = −1; absolute minimum −4 at x = 8.
Both answers sit at endpoints, even though the function has a local minimum at x = 0 and a local maximum at x = 1. A student who finds only where f′ = 0 checks one point and gets both answers wrong.
For optimisation word problems and related rates, see our post on applications of maxima and minima.
Which mistakes cost the most marks?
Skipping the endpoints. This is the slip the whole method is built to prevent.
Missing points where f′ does not exist, such as x = 0 for x^(2/3) or the corner of |x − a|.
Keeping critical points outside the interval.
Reporting the x-value when the question asks for the maximum value, or the other way round.
Assuming a local maximum is the absolute maximum. Compare it with the endpoints first.
A practice routine
Dunlosky and colleagues (2013) rated interleaved practice, mixing problem types, as moderate utility and practice testing as high utility. Mix the three types above rather than doing ten of one kind.
Candidate list drill: for eight functions, write only the list of points to check. Do not evaluate yet.
Mixed set: six functions, two of each type above (critical point outside, answer at an endpoint, f′ undefined), shuffled.
Sketch check: for two of them, sketch the graph and mark the absolute maximum and minimum.
If you keep missing a candidate point and cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find the absolute extrema of f(x) = |x − 1| + x on [−2, 3], and list the candidates before evaluating.
