Question
Find the maximum and minimum values of on the closed interval . A student finds critical points at and , evaluates at those, and reports as max and as min. Where did they go wrong?
Solution — Step by Step
Critical points: and . Both lie inside .
This is where the student went wrong. On a closed interval, we always check critical points and both endpoints.
.
(local max).
(local min).
.
Values: 24, 29, 28, 56. Maximum is at the right endpoint, not at the critical point!
Maximum value = . Minimum value = .
The student missed both. They reported a local max at 29 instead of the global max at 56, and a local min at 28 instead of the global min at 24.
Why This Works
The Extreme Value Theorem guarantees that a continuous function on a closed bounded interval attains its absolute max and min. Those extrema occur at either (a) interior critical points where or undefined, or (b) the endpoints. So the search list always has both kinds.
A local max/min from the derivative test is not the same as the absolute max/min on the interval. Endpoints can easily beat interior critical points — especially for cubic and higher-degree polynomials whose end behaviour shoots up.
Alternative Method
Sketch the function. is a cubic with positive leading coefficient. has roots at 2 and 3, with on and on . So is increasing-decreasing-increasing. On : increasing from to , decreasing to , then increasing again to . Sketch confirms 56 is the absolute max.
Common Mistake
The error in the question is the most common one in JEE Main on AOD: forgetting endpoints on a closed interval. Always make a list with evaluated at every critical point plus both endpoints. Pick the largest and smallest — those are the answers. This single habit fixes 90% of careless mistakes in optimisation problems.