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JEE Main 2022 · Mathematics · Differentiation and Application of Derivatives · medium

The minimum value of f(x)=x^3-12x+5 on the interval [0,3] is

Single correct option

The minimum value of f(x)=x3−12x+5f(x)=x^3-12x+5 on the interval [0,3][0,3] is
  1. A. 55
  2. B. −4-4
  3. C. −11-11
  4. D. −16-16
Show answer and solution

Answer: (C) −11-11

f′(x)=3x2−12=0f'(x)=3x^2-12=0 gives x=2x=2 in the interval. Evaluate at the critical point and endpoints: f(0)=5f(0)=5, f(2)=8−24+5=−11f(2)=8-24+5=-11, f(3)=27−36+5=−4f(3)=27-36+5=-4. The least value is −11-11. The value −16-16 arises from using the critical point x=−2x=-2, which lies outside the interval.
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