Application of Derivatives: PYQ Walkthrough (4)

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Question

Find the maximum value of f(x)=2x315x2+36x7f(x) = 2x^3 - 15x^2 + 36x - 7 on the interval [1,5][1, 5]. (CBSE 2023 4-mark PYQ)

Solution — Step by Step

f(x)=6x230x+36=6(x25x+6)=6(x2)(x3)f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)

Critical points: x=2x = 2 and x=3x = 3. Both lie inside the interval [1,5][1, 5], so they’re candidates for the maximum.

f(1)=215+367=16f(1) = 2 - 15 + 36 - 7 = 16 f(2)=1660+727=21f(2) = 16 - 60 + 72 - 7 = 21 f(3)=54135+1087=20f(3) = 54 - 135 + 108 - 7 = 20 f(5)=250375+1807=48f(5) = 250 - 375 + 180 - 7 = 48

Among {16,21,20,48}\{16, 21, 20, 48\}, the maximum is 4848 at x=5x = 5.

Final answer: Maximum value is 4848, attained at x=5x = 5.

Why This Works

For continuous functions on a closed interval, the absolute extrema occur either at critical points (where f(x)=0f'(x) = 0) or at the endpoints. Always check all four candidates — never assume the critical point gives the maximum.

CBSE board examiners deliberately design intervals where the maximum is at an endpoint, not at a critical point. Students who skip endpoints lose 2 marks.

Alternative Method

Test the second derivative at each critical point: f(x)=12x30f''(x) = 12x - 30. f(2)=6<0f''(2) = -6 < 0 (local max), f(3)=6>0f''(3) = 6 > 0 (local min). Local max at x=2x=2 gives 2121, but absolute max is still found by comparing with endpoints.

The closed-interval method works in 3 steps: find ff', list candidates (f=0f'=0 inside + endpoints), evaluate ff at each. Don’t overthink — JEE Main 2024 had this exact pattern.

Common Mistake

Stopping at the local maximum x=2x=2 and reporting f(2)=21f(2) = 21 as the answer. The interval extends to x=5x=5, where ff is much larger. Always evaluate at endpoints.

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