at what x value does f(x) have an absolute (global) maximum and minimum on -2,3? absolute maximum x =…

Explanation:

Step1: Recall critical - point concept

Critical points occur where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined. From the graph of $f^{\prime}(x)$, $f^{\prime}(x) = 0$ at $x=-2,0,3$.

Step2: Analyze the behavior of $f(x)$ using $f^{\prime}(x)$

• For $x\in[-2,0)$, $f^{\prime}(x)>0$, so $f(x)$ is increasing.
• For $x\in(0,3]$, $f^{\prime}(x)<0$, so $f(x)$ is decreasing.

Step3: Evaluate $f(x)$ at critical and endpoints

We evaluate $f(x)$ at the critical points $x = - 2,0,3$ (the endpoints of the interval $[-2,3]$ are also critical - like in the context of the closed - interval method). Since $f(x)$ is increasing on $[-2,0)$ and decreasing on $(0,3]$, the absolute maximum occurs at $x = 0$ and the absolute minimum occurs at either $x=-2$ or $x = 3$. To be sure, we note that the function changes from increasing to decreasing at $x = 0$.

Answer:

Absolute Maximum: $x = 0$ Absolute Minimum: $x=-2$ or $x = 3$ (more information about the function values at these points would be needed to determine exactly which one is the minimum, but based on the derivative behavior, these are the candidates)