at what x value does f(x) have an absolute (global) maximum and minimum on -3.5, 3? absolute maximum x =…

Explanation:

Step1: Recall the relationship between $f(x)$ and $f'(x)$

Critical - points of $f(x)$ occur where $f'(x)=0$ or $f'(x)$ is undefined. From the graph of $f'(x)$, $f'(x) = 0$ at $x=- 1,x = 2$. Also, we need to consider the endpoints of the interval $[-3.5,3]$, which are $x=-3.5$ and $x = 3$.

Step2: Analyze the behavior of $f(x)$ using $f'(x)$

When $f'(x)>0$, $f(x)$ is increasing. When $f'(x)<0$, $f(x)$ is decreasing.

• For $x\in[-3.5,-1]$, $f'(x)<0$, so $f(x)$ is decreasing.
• For $x\in[-1,2]$, $f'(x)>0$, so $f(x)$ is increasing.
• For $x\in[2,3]$, $f'(x)<0$, so $f(x)$ is decreasing.

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

We know that $f(x)$ has a local minimum at $x=-1$ (since $f(x)$ changes from decreasing to increasing) and a local maximum at $x = 2$ (since $f(x)$ changes from increasing to decreasing). We also need to evaluate $f(x)$ at the endpoints $x=-3.5$ and $x = 3$. By comparing the values of $f(x)$ at $x=-3.5,x=-1,x = 2,x = 3$, we find that: The absolute maximum of $f(x)$ on $[-3.5,3]$ occurs at $x = 2$. The absolute minimum of $f(x)$ on $[-3.5,3]$ occurs at $x=-1$.

Answer:

Absolute Maximum: $x = 2$ Absolute Minimum: $x=-1$