Optimization on a Closed Interval

Suppose you’ve found the peaks and valleys in the interior of an interval. Are you done? Not yet — the exam usually wants the single highest and lowest value f takes anywhere on [a, b], and that winner can easily sit at one of the endpoints. A local peak in the middle is not automatically the tallest point in the room.

There’s a clean theorem that makes the search short. If f is continuous on a closed interval [a, b], the Extreme Value Theorem promises that f actually attains a global max and a global min somewhere on the interval. And those winners can only live in two places: at an interior critical point, or at an endpoint. That’s a short, finite list of suspects — exactly what GATE wants you to check.

This “check the interior and the boundary” habit is exactly what you do when tuning a hyperparameter over a fixed range (a learning rate in [0.001, 0.1], say): the best value is sometimes inside, but just as often it’s pinned at one end.

The closed-interval method

To find the global max and min of a continuous f on [a, b]:

1.  Find the critical points inside (a, b):  solve f'(x) = 0
    (and note any x in (a, b) where f' is undefined).
2.  List the candidates:  those interior critical points  +  the endpoints a and b.
3.  Evaluate f at EVERY candidate.
4.  Largest value = global maximum.   Smallest value = global minimum.

No second-derivative test is needed — you are not classifying the points, just comparing their heights. The whole method is “evaluate and compare.”

Play with this on a cubic. Pick x³ − x and slide the point across the plotted domain. The tangent flattens at the two interior critical points — but watch what happens at the ends of the curve: the function climbs higher (and dips lower) there than at either critical point. On a closed interval those endpoint values are candidates too, and the global winner is often one of them.

How GATE asks this

Look for a NAT or MSQ that says “find the global maximum (or minimum) value of f on [a, b],” or an MSQ probing why you must check endpoints or when the Extreme Value Theorem applies. The reliable trap: a candidate forgets the endpoints and reports an interior local extremum as the global answer. GATE DA 2025 used this shape — a polynomial on a closed interval where one endpoint held the true extreme.

Worked example — based on a real GATE DA 2025 question

Find the global maximum and global minimum of f(x) = x³ − 3x on the closed interval [−2, 2].

Step 1 — interior critical points. Same derivative as before:

f'(x) = 3x² − 3 = 0   →   x = −1  and  x = +1     (both lie inside (−2, 2))

Step 2-3 — evaluate f at all four candidates (two critical points, two endpoints):

candidate     type              f(x) = x³ − 3x
──────────────────────────────────────────────────────
x = −2        endpoint          (−2)³ − 3(−2) = −8 + 6 = −2
x = −1        critical point    (−1)³ − 3(−1) = −1 + 3 =  2
x = +1        critical point    (1)³  − 3(1)  =  1 − 3 = −2
x = +2        endpoint          (2)³  − 3(2)  =  8 − 6 =  2

Step 4 — compare. The largest value is 2; the smallest is −2:

global maximum = 2   (attained at x = −1 and at x = 2)
global minimum = −2  (attained at x = 1 and at x = −2)

Notice that the global maximum is tied between an interior critical point (x = −1) and an endpoint (x = 2), and the global minimum is tied between x = 1 and the endpoint x = −2. Had you skipped the endpoints, you would have missed half of where the extremes actually occur — and on a different interval the endpoint could have been the sole winner.

Quick check