consider the function ( f(x)=1 - 6x^{2},-4leq xleq2). the absolute maximum value is and this occurs at ( x =…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=1 - 6x^{2}$ using the power - rule. The derivative $f^\prime(x)=\frac{d}{dx}(1)-6\frac{d}{dx}(x^{2})=0 - 12x=-12x$.

Step2: Find critical points

Set $f^\prime(x) = 0$. So, $-12x = 0$, which gives $x = 0$.

Step3: Evaluate the function at critical and endpoints

Evaluate $f(x)$ at $x=-4,x = 0,x = 2$. When $x=-4$, $f(-4)=1-6\times(-4)^{2}=1 - 96=-95$. When $x = 0$, $f(0)=1-6\times0^{2}=1$. When $x = 2$, $f(2)=1-6\times2^{2}=1 - 24=-23$.

Answer:

The absolute maximum value is $1$ and this occurs at $x = 0$. The absolute minimum value is $-95$ and this occurs at $x=-4$.