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

Explanation:

Step1: Find the derivative

Differentiate $f(x)=2x^{3}-x^{2}-4x + 6$ using the power - rule. $f^\prime(x)=6x^{2}-2x - 4$.

Step2: Set the derivative equal to zero

Solve $6x^{2}-2x - 4 = 0$. Factor out a 2: $2(3x^{2}-x - 2)=0$. Then factor the quadratic: $2(3x + 2)(x - 1)=0$. So $x=-\frac{2}{3}$ or $x = 1$. But $x = 1$ is not in the interval $[-1,0]$, so we discard it.

Step3: Evaluate the function at critical and endpoints

Evaluate $f(x)$ at $x=-1$, $x=-\frac{2}{3}$, and $x = 0$. $f(-1)=2(-1)^{3}-(-1)^{2}-4(-1)+6=-2 - 1+4 + 6=7$. $f(-\frac{2}{3})=2(-\frac{2}{3})^{3}-(-\frac{2}{3})^{2}-4(-\frac{2}{3})+6=2(-\frac{8}{27})-\frac{4}{9}+\frac{8}{3}+6=-\frac{16}{27}-\frac{12}{27}+\frac{72}{27}+6=\frac{-16 - 12+72}{27}+6=\frac{44}{27}+6=\frac{44 + 162}{27}=\frac{206}{27}\approx7.63$. $f(0)=2(0)^{3}-(0)^{2}-4(0)+6=6$.

Answer:

The absolute maximum value is $\frac{206}{27}$ at $x=-\frac{2}{3}$.