Q.Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

We find absolute extrema on a closed interval by evaluating the function at critical points (where $f'(x)=0$ or $f'$ does not exist) and at the endpoints.

(i) $f(x)=x^3$, $x\in[-2,2]$

$f'(x)=3x^2=0 \Rightarrow x=0$ (critical point).

$f(-2)=-8$, $f(0)=0$, $f(2)=8$. …

(i) $f(x) = x^3$ on $[-2, 2]$

Why this works: The function $x^3$ is strictly increasing everywhere (its derivative $3x^2 \ge 0$ and zero only at $x=0$). On a closed interval, a strictly increasing function attains its minimum at the left endpoint and its maximum at the right endpoint. No critical point inside can beat the endpoints.

1. Find critical points: $f'(x) = 3x^2 = 0 \implies x = 0$. This is inside $[-2, 2]$.
2. Evaluate at endpoints and critical point:
   • $f(-2) = (-2)^3 = -8$
   • $f(0) = 0$
   • $f(2) = 8$
3. Compare: The smallest value is $-8$ at $x = -2$; the largest is $8$ at $x = 2$. …

Method: The Closed-Interval (Candidates) Test for Absolute Extrema

Whenever a question restricts a continuous function to a closed interval $[a,b]$ and asks for the absolute maximum/minimum, this exact procedure applies regardless of whether the function is a polynomial, trigonometric, or otherwise.

Steps

Step 1: Confirm the function is continuous on the closed interval.

A continuous function on a closed, bounded interval is guaranteed to attain both an absolute maximum and an absolute minimum somewhere in that interval, so the search is never in vain.

Step 2: Find every critical point of $f$ that lies strictly inside $(a,b)$.

Solve $f'(x)=0$ (and note any point where $f'$ fails to exist), then keep only the solutions that fall inside the open interval — critical points outside $[a,b]$ are irrelevant.

Step 3: List the "candidates": the interior critical points together with the two endpoints $a$ and $b$.

$\text{candidates} = \{a, b\} \cup \{\, x : f'(x)=0,\ a<x<b \,\}$ …

Common Mistakes

Mistake 1: Forgetting to include the endpoints as candidates.

Why it's wrong: in part (iii), the absolute minimum $-10$ actually occurs at the left endpoint $x=-2$, not at the interior critical point $x=4$ — a student who only checks $f'(x)=0$ and never plugs in the endpoints would miss it entirely. Correct approach: always evaluate $f$ at both endpoints of the closed interval in addition to every interior critical point, with no exceptions.

Mistake 2: Assuming the interior critical point is automatically the maximum (or minimum) without comparing.

Why it's wrong: in part (i), $x=0$ is the only critical point of $f(x)=x^3$, yet it is neither the absolute max nor the absolute min on $[-2,2]$ — both extremes occur at the endpoints instead, since $x=0$ is only a point of inflection. Correct approach: never skip the comparison step; a critical point on a closed interval is just one more candidate, not an automatic winner. …