find the absolute maximum and absolute minimum values of f on the given interval.\n$f(x)=2x^{3}-3x^{2}-72x +…

find the absolute maximum and absolute minimum values of f on the given interval.\n$f(x)=2x^{3}-3x^{2}-72x + 8$, $-4,5$

find the absolute maximum and absolute minimum values of f on the given interval.\n$f(x)=2x^{3}-3x^{2}-72x + 8$, $-4,5$

Answer

Explanation:

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

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (f(x)=2x^{3}-3x^{2}-72x + 8), we have (f^\prime(x)=6x^{2}-6x - 72).

Step2: Find the critical points

Set (f^\prime(x)=0), so (6x^{2}-6x - 72 = 0). Divide through by (6) to get (x^{2}-x - 12=0). Factor the quadratic: ((x - 4)(x+ 3)=0). Solving (x - 4=0) gives (x = 4), and solving (x + 3=0) gives (x=-3). Both (x=-3) and (x = 4) are in the interval ([-4,5]).

Step3: Evaluate (f(x)) at critical points and endpoints

  • For (x=-4): (f(-4)=2(-4)^{3}-3(-4)^{2}-72(-4)+8=2(-64)-3(16)+288 + 8=-128-48 + 288+8=128).
  • For (x=-3): (f(-3)=2(-3)^{3}-3(-3)^{2}-72(-3)+8=2(-27)-3(9)+216 + 8=-54-27+216 + 8=143).
  • For (x = 4): (f(4)=2(4)^{3}-3(4)^{2}-72(4)+8=2(64)-3(16)-288 + 8=128-48-288 + 8=-200).
  • For (x = 5): (f(5)=2(5)^{3}-3(5)^{2}-72(5)+8=2(125)-3(25)-360 + 8=250-75-360 + 8=-177).

Answer:

The absolute maximum value of (f(x)) on the interval ([-4,5]) is (143) (at (x=-3)), and the absolute minimum value is (-200) (at (x = 4)).