consider the function $f(x)=2 - 7x^{2}, -3leq xleq1$. the absolute maximum value is and this occurs at $x$…

Explanation:

Step1: Find the derivative

The derivative of $f(x)=2 - 7x^{2}$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=-14x$.

Step2: Find critical points

Set $f'(x) = 0$. So, $-14x=0$, which gives $x = 0$.

Step3: Evaluate the function at critical and endpoints

Evaluate $f(x)$ at $x=-3,x = 0,x = 1$. When $x=-3$, $f(-3)=2-7\times(-3)^{2}=2 - 63=-61$. When $x = 0$, $f(0)=2-7\times0^{2}=2$. When $x = 1$, $f(1)=2-7\times1^{2}=2 - 7=-5$.

Answer:

The absolute maximum value is $2$ and this occurs at $x = 0$. The absolute minimum value is $-61$ and this occurs at $x=-3$.