ExamBro
ExamBro
AP EAMCET · Maths · Application of Derivatives

The difference between the absolute maximum and absolute minimum values of the function \(f(x)=2 x^3-15 x^2+36 x-30\) on \([-1,4]\) is

  1. A \(80\)
  2. B \(1\)
  3. C \(85\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(85\)

Step-by-step Solution

Detailed explanation

\(f'(x) = 6x^2 - 30x + 36\) \(6x^2 - 30x + 36 = 0 \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0 \Rightarrow x=2,3\) \(f(-1) = 2(-1)^3 - 15(-1)^2 + 36(-1) - 30 = -83\) \(f(2) = 2(2)^3 - 15(2)^2 + 36(2) - 30 = -2\) \(f(3) = 2(3)^3 - 15(3)^2 + 36(3) - 30 = -3\)…