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

Explanation:

Step1: Find the derivative

$f'(x)=\frac{d}{dx}(8 + 3x-3x^{2})=3 - 6x$

Step2: Find critical points

Set $f'(x)=0$, so $3 - 6x = 0$. Solving for $x$ gives $x=\frac{1}{2}$.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x = 0$, $x=\frac{1}{2}$, and $x = 4$. $f(0)=8+3\times0 - 3\times0^{2}=8$ $f(\frac{1}{2})=8+3\times\frac{1}{2}-3\times(\frac{1}{2})^{2}=8+\frac{3}{2}-\frac{3}{4}=\frac{32 + 6-3}{4}=\frac{35}{4}=8.75$ $f(4)=8+3\times4-3\times4^{2}=8 + 12-48=-28$

Answer:

The absolute maximum value is $8.75$ and it occurs at $x=\frac{1}{2}$. The absolute minimum value is $-28$ and it occurs at $x = 4$.