consider the function $f(x)=2x^{3}+12x^{2}-30x + 4$, $-5leq xleq2$. this function has an absolute minimum…

consider the function $f(x)=2x^{3}+12x^{2}-30x + 4$, $-5leq xleq2$. this function has an absolute minimum value equal to and an absolute maximum value equal to

consider the function $f(x)=2x^{3}+12x^{2}-30x + 4$, $-5leq xleq2$. this function has an absolute minimum value equal to and an absolute maximum value equal to

Answer

Explanation:

Step1: Find the derivative of (f(x))

The derivative (f^\prime(x)) of (f(x)=2x^{3}+12x^{2}-30x + 4) is (f^\prime(x)=6x^{2}+24x - 30). Factor out (6): (f^\prime(x)=6(x^{2}+4x - 5)=6(x + 5)(x - 1)).

Step2: Find the critical points

Set (f^\prime(x)=0). Using the zero - product property (6(x + 5)(x - 1)=0), we get (x=-5) or (x = 1). Both (x=-5) and (x = 1) are in the interval ([-5,2]).

Step3: Evaluate the function at the critical points and endpoints

  • For (x=-5): (f(-5)=2(-5)^{3}+12(-5)^{2}-30(-5)+4=2(-125)+12(25)+150 + 4=-250+300+150 + 4=204).
  • For (x = 1): (f(1)=2(1)^{3}+12(1)^{2}-30(1)+4=2 + 12-30 + 4=-12).
  • For (x = 2): (f(2)=2(2)^{3}+12(2)^{2}-30(2)+4=2(8)+12(4)-60 + 4=16+48-60 + 4=8).

Answer:

The absolute minimum value is (-12) and the absolute maximum value is (204).