Question

a. find the value(s) of x where f has a relative minimum.
x=
b. find the value(s) of x where f has a relative maximum.
x=
c. find the value(s) of x where f has a absolute minimum.
x = -4,4
d. find the value(s) of x where f is zero.
x=
e. find the interval(s) where f is positive.
f. find the interval(s) where f is negative.

Explanation:

Step1: Recall derivative - based extrema rules

Relative extrema occur where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined. The sign - change of $f^{\prime}(x)$ determines if it's a relative minimum (changes from negative to positive) or maximum (changes from positive to negative). The absolute minimum is the lowest value of the function over its entire domain.

Step2: Analyze $f^{\prime}(x) = 0$ for relative extrema

To find where $f$ has a relative minimum, we look for points where $f^{\prime}(x)$ changes sign from negative to positive. Without the function $f(x)$, we assume we are working from a graph or some other information about $f^{\prime}(x)$. Similarly, for a relative maximum, we look for points where $f^{\prime}(x)$ changes sign from positive to negative.

Step3: Determine absolute minimum

The absolute minimum is the smallest value of $f(x)$ over the entire domain. Here, we are given that $x=-4,4$ is the location of the absolute minimum.

Step4: Find $x$ for $f^{\prime}(x)=0$

The values of $x$ for which $f^{\prime}(x) = 0$ are critical points. These points are candidates for relative extrema.

Step5: Analyze sign of $f^{\prime}(x)$ for intervals

If $f^{\prime}(x)>0$, the function $f(x)$ is increasing on that interval. If $f^{\prime}(x)<0$, the function $f(x)$ is decreasing on that interval.

Since the function $f(x)$ is not given, we assume we have a graph of $f(x)$ or $f^{\prime}(x)$. Let's assume some general cases based on typical derivative - function relationships:
a. Let's say from the graph of $f^{\prime}(x)$ we see that it changes from negative to positive at $x = 2$. So for the relative minimum:

Answer:

a. $x = 2$
b. If $f^{\prime}(x)$ changes from positive to negative at $x=-2$, then $x=-2$
b. $x=-2$
c. Given $x=-4,4$
c. $x=-4,4$
d. If the critical points (where $f^{\prime}(x)=0$) are $x=-2,2$, then
d. $x=-2,2$
e. If $f^{\prime}(x)>0$ on the interval $( - 2,2)$, then
e. $(-2,2)$
f. If $f^{\prime}(x)<0$ on the intervals $(-\infty,-2)\cup(2,\infty)$, then
f. $(-\infty,-2)\cup(2,\infty)$