Question

step 1
the absolute maximum and minimum values of f occur either at a critical point inside the interval or at an endpoint of the interval. recall that a critical point is a point where f(x)=0 or is undefined. we begin by finding the derivative of f.
f(x)=18x^{2}-36x - 54

step 2
we now solve f(x)=0 for x, which gives the following critical numbers. (enter your answers as a comma - separated list.)
x=-1,3

step 3
we must now find the function values at the critical numbers we just found and at the endpoints of the interval -2,4.
f(-1)=38
f(3)=-154
f(-2)=4
f(4)=-112

step 4
therefore, on the interval -2,4, the absolute minimum value of f(x) is and the absolute maximum value is

Explanation:

Step1: Find derivative

Given the derivative $f'(x)=18x^{2}-36x - 54$. Critical points are where $f'(x) = 0$.

Step2: Solve for critical points

Set $18x^{2}-36x - 54=0$. Divide through by 18: $x^{2}-2x - 3=0$. Factor to $(x + 1)(x - 3)=0$. So $x=-1,3$.

Step3: Evaluate function at critical and end - points

Evaluate $f(x)$ at $x=-1,3,-2,4$. We have $f(-1) = 38$, $f(3)=-154$, $f(-2)=4$, $f(4)=-112$.

Step4: Determine max and min

Compare the function values. The smallest value among $38,-154,4,-112$ is $-154$ and the largest is $38$.

Answer:

The absolute minimum value of $f(x)$ is $-154$ and the absolute maximum value is $38$.