find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x…

Explanation:

Step1: Find the derivative

$f'(x)=6x^{2}-4x - 2$

Step2: Set the derivative equal to 0

$6x^{2}-4x - 2 = 0$ Factor out 2: $2(3x^{2}-2x - 1)=0$ Factor the quadratic: $2(3x + 1)(x - 1)=0$ Solve for $x$: $x=-\frac{1}{3},1$

Step3: Evaluate the function at critical - points and endpoints

Evaluate $f(x)$ at $x=-1,-\frac{1}{3},1,2$ $f(-1)=2(-1)^{3}-2(-1)^{2}-2(-1)+7=-2 - 2 + 2+7=5$ $f(-\frac{1}{3})=2(-\frac{1}{3})^{3}-2(-\frac{1}{3})^{2}-2(-\frac{1}{3})+7=2(-\frac{1}{27})-2(\frac{1}{9})+\frac{2}{3}+7=-\frac{2}{27}-\frac{2}{9}+\frac{2}{3}+7=-\frac{2}{27}-\frac{6}{27}+\frac{18}{27}+7=\frac{-2 - 6+18}{27}+7=\frac{10}{27}+7 = 7\frac{10}{27}\approx7.37$ $f(1)=2(1)^{3}-2(1)^{2}-2(1)+7=2 - 2 - 2+7=5$ $f(2)=2(2)^{3}-2(2)^{2}-2(2)+7=16 - 8 - 4+7=11$

Answer:

The absolute maximum value is 11, which occurs at $x = 2$. The absolute minimum value is 5, which occurs at $x=-1$ and $x = 1$.