Q7
1 markMCQSection A

The least value of f(x)=x3−12x, x∈[0,3]f(x) = x^3 - 12x,\ x \in [0, 3] is

Application of Derivatives
Maxima and Minima on a Closed Interval

Options

(A)-16
(B)-9
(C)0
(D)16
Official Answer

Correct option: (A) -16


Setting f′(x)=3x2−12=0 gives x=2 (within [0,3])f'(x) = 3x^2 - 12 = 0 \text{ gives } x = 2\ (\text{within } [0,3]). Evaluating f at the critical point and endpoints: f(0)=0, f(2)=−16, f(3)=−9f(0) = 0,\ f(2) = -16,\ f(3) = -9. The least value is -16.

absolute minimum closed intervalcritical pointsfirst derivative testmaxima and minimax^3-12xendpoint evaluation

Marking Scheme

  • 11 mark: correct option (A) selected — requires finding the critical point and comparing with both endpoint values.

Hint

For absolute extrema on a closed interval, always evaluate f at every critical point inside the interval AND at both endpoints, then compare all values.

Quick Oral Answer

The critical point in [0,3][0,3] is x=2 where f(2)=−16x=2 \text{ where } f(2)=-16, which is lower than both endpoint values f(0)=0 and f(3)=−9f(0)=0 \text{ and } f(3)=-9, so the least value is -16, option (A).

Analysis & Explanation

This is a standard closed-interval absolute extrema problem requiring evaluation at both critical points and endpoints.


Finding critical points

  • f(x)=x3−12x, so f′(x)=3x2−12f(x) = x^3 - 12x, \text{ so } f'(x) = 3x^2 - 12.
  • Setting f′(x)=0:3x2=12  ⟹  x2=4  ⟹  x=±2f'(x) = 0: 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2. Only x=2 lies in [0,3] (x=−2 is rejected as outside the interval)x = 2 \text{ lies in } [0, 3]\ (x = -2 \text{ is rejected as outside the interval}).

Evaluating at critical point and endpoints

  • f(0)=0−0=0f(0) = 0 - 0 = 0
  • f(2)=8−24=−16f(2) = 8 - 24 = -16
  • f(3)=27−36=−9f(3) = 27 - 36 = -9

Identifying the least value

  • Comparing 0, -16, -9: the smallest is -16, occurring at x = 2. This matches option (A).

Why the other options are wrong

  • (B) -9 is the value at the endpoint x=3x = 3, not the overall minimum.
  • (C) 0 is the value at the other endpoint x=0x = 0, again not the minimum.
  • (D) 16 does not correspond to any evaluated point and is simply the positive counterpart of the correct answer, a likely sign-error distractor.

Common Mistakes

  1. 1Forgetting to check the interval boundary and rejecting x=2x = 2 mistakenly, or including x=−2x = -2 which lies outside [0,3][0,3].
  2. 2Only checking the critical point value (−16)(-16) or only the endpoints, without comparing all three values to find the true least value.
  3. 3Sign errors while computing f(2)=23−12(2)f(2) = 2^3 - 12(2), leading to +16+16 instead of −16-16.

Interesting Facts

This method — checking critical points and both endpoints — is called the Closed Interval Method and is guaranteed to work by the Extreme Value Theorem, since a continuous function always attains its absolute max and min on a closed, bounded interval.

The function x3−12xx^3 - 12x is a classic cubic with local extrema at x=±2x = \pm 2; on an unrestricted domain it has no absolute minimum or maximum since it tends to −∞-\infty and +∞+\infty at the ends.

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Frequently Asked Questions

Why must both endpoints be checked along with critical points?

On a closed interval, the absolute maximum or minimum of a continuous function can occur either at an interior critical point (where f′(x)=0f'(x)=0 or is undefined) or at one of the endpoints. Skipping endpoint evaluation can miss the true extreme value, as happens here where the endpoint x=3 gives −9x=3 \text{ gives } -9, close to but not equal to the true minimum.

Why is x=−2x = -2 rejected as a critical point here?

The domain restriction is x∈[0,3]x \in [0, 3]. Since -2 does not lie within this closed interval, it is not a valid critical point to consider for this particular problem, even though it satisfies f′(x)=0f'(x) = 0 mathematically.