Q.The smallest value of the polynomial $x^3 - 18x^2 + 96x$ in $[0, 9]$ is:
(A) $126$
(B) $0$
(C) $135$
(D) $160$

Concept: Quadratic Extrema — The cubic reduces to a quadratic after differentiation; we find critical points and check endpoints.

Step 1: Differentiate

$f(x) = x^3 - 18x^2 + 96x$

$f'(x) = 3x^2 - 36x + 96 = 3(x^2 - 12x + 32)$

Step 2: Find critical points

$x^2 - 12x + 32 = 0 \implies (x-4)(x-8)=0$

So $x = 4$ and $x = 8$ lie inside $[0,9]$.

Step 3: Evaluate $f$ at endpoints and critical points

$f(0) = 0$ …

We are asked for the minimum of $f(x) = x^3 - 18x^2 + 96x$ on the closed interval $[0,9]$. Since this is a continuous function on a closed interval, the extreme value theorem guarantees that the minimum occurs either at a critical point inside the interval or at one of the endpoints.

Why this approach? For a polynomial, the derivative tells us where the slope is zero — those are the candidate peaks and valleys. But a cubic can have both a local max and a local min, so we must check which of these (if any) lies inside $[0,9]$, and then compare their values with the endpoints.

1. Find the derivative and critical points

   $f'(x) = 3x^2 - 36x + 96$

   Set $f'(x) = 0$:

Divide through by 3:

Factor:

So $x = 4$ and $x = 8$ are the critical points. Both lie inside $[0,9]$.

2. Classify each critical point (optional but clarifying)

   The second derivative: $f''(x) = 6x - 36$

   • At $x = 4$: $f''(4) = 24 - 36 = -12 < 0$ → local maximum
   • At $x = 8$: $f''(8) = 48 - 36 = 12 > 0$ → local minimum

   So the cubic rises to a peak at $x=4$, then falls to a trough at $x=8$, then rises again.

3. Evaluate $f$ at all candidates

   We need the values at $x = 0$, $x = 4$, $x = 8$, and $x = 9$:

   • $f(0) = 0^3 - 18(0)^2 + 96(0) = 0$
   • $f(4) = 64 - 18(16) + 96(4) = 64 - 288 + 384 = 160$
   • $f(8) = 512 - 18(64) + 96(8) = 512 - 1152 + 768 = 128$ …

Method: Finding Absolute (Global) Extrema on a Closed Interval

Use this whenever a question asks for the largest or smallest value a continuous function takes over a closed interval $[a,b]$.

Steps

Step 1: Differentiate the function and find all critical points.

Solve $f'(x)=0$ (and note any points where $f'$ fails to exist but $f$ is still defined) to get the candidate interior points.

Step 2: Keep only the critical points that lie inside the given interval.

Discard any critical point falling outside $[a,b]$ — it cannot affect the extreme values on this restricted domain.

Step 3: Evaluate $f$ at every surviving critical point AND at both endpoints $a$ and $b$. …

Common Mistakes

Mistake 1: Stopping after finding the local minimum from the critical points, without checking the endpoints.

Why it's wrong: the absolute smallest value on a closed interval can occur at an endpoint even when it isn't a stationary point at all — here the true minimum ($0$, at $x=0$) is smaller than the local minimum found from the derivative ($128$, at $x=8$). Correct approach: always evaluate $f$ at both endpoints of the interval in addition to every interior critical point, then compare all the values together.

Mistake 2: Mishandling critical points that fall outside the interval.

Why it's wrong: a critical point outside $[0,9]$ has no bearing on the extreme values within that restricted domain, and including it wrongly leads to comparing the wrong set of numbers. Correct approach: solve $f'(x)=0$ first, then explicitly check which roots actually lie inside the given closed interval before evaluating anything. …