Q.Find the maximum value of $2x^3-24x+107$ in the interval $[1, 3]$. Find the maximum value of the same function in $[-3, -1]$.

Idea: on each closed interval, compare $f$ at its interior critical points and its endpoints.

Let $f(x)=2x^3-24x+107$, so

giving critical points $x=2$ and $x=-2$.

On $[1,3]$ (only $x=2$ is inside):

Largest $=89$ at $x=3$.

On $[-3,-1]$ (only $x=-2$ is inside): …

The plan

$f$ is a polynomial, hence continuous on each closed interval, so it attains a maximum there. That maximum sits either at an interior critical point (where $f'(x)=0$) or at an endpoint. So we find the critical points once, then for each interval evaluate $f$ at the critical point that falls inside it and at the two endpoints, and keep the largest value.

Step 1 — Critical points

Setting $f'(x)=0$ gives $x=2$ and $x=-2$.

Step 2 — Interval $[1,3]$

Only $x=2$ lies inside $[1,3]$, so the candidates are $x=1,2,3$:

• $f(1)=2-24+107=85,$
• $f(2)=16-48+107=75,$
• $f(3)=54-72+107=89.$

The largest is $89$, at the endpoint $x=3$.

Step 3 — Interval $[-3,-1]$ …

Method: Comparing the Same Function's Maximum Across Different Intervals

This teaches the closed-interval candidates test applied twice to the same function on two different closed intervals — the key skill is realising the critical points don't change, but which ones matter (and which endpoints matter) does.

Steps

Step 1: Find all critical points of the function once, ignoring intervals for now.

Differentiate $f(x)$ and solve $f'(x)=0$ over all of $\mathbb{R}$ — for a cubic like $2x^3-24x+107$, this typically gives two critical points (a local max and a local min candidate).

Step 2: For each interval separately, keep only the critical point(s) that actually lie inside it.

Different intervals will "see" different critical points — an interval like $[1,3]$ only contains $x=2$, while $[-3,-1]$ only contains $x=-2$; never carry a critical point into an interval it doesn't belong to. …

Common Mistakes

Mistake 1: Using the same critical point for both intervals.

Why it's wrong: $x=2$ is the only critical point relevant to $[1,3]$, and $x=-2$ is the only one relevant to $[-3,-1]$ — a student who reuses $x=2$ when evaluating the second interval is checking a point that isn't even inside $[-3,-1]$. Correct approach: for each interval, explicitly re-check which of the function's critical points actually fall within that interval's bounds before building the candidate list.

Mistake 2: Assuming the maximum must be at the same "type" of location (endpoint vs. critical point) in both intervals.

Why it's wrong: here the maximum on $[1,3]$ occurs at an endpoint ($x=3$), while the maximum on $[-3,-1]$ occurs at the interior critical point ($x=-2$) — assuming both intervals behave the same way skips the actual comparison. Correct approach: evaluate and compare every candidate freshly for each interval rather than pattern-matching from the first interval's result. …