Question

answer the following questions about the function whose derivative is f(x)=(x + 5)e^{-2x}.
a. what are the critical points of f?
b. on what open intervals is f increasing or decreasing?
c. at what points, if any, does f assume local maximum or minimum values?
a. what are the critical points of f? select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. the critical point(s) of f is/are x =
(simplify your answer. use a comma to separate answers as needed.)
b. the function f has no critical points.

Explanation:

Step1: Recall critical - point definition

Critical points occur where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined. Since $y = e^{-2x}=\frac{1}{e^{2x}}$ is never zero and is defined for all real $x$, we set $f^{\prime}(x)=(x + 5)e^{-2x}=0$.

Step2: Solve for $x$

Since $e^{-2x}
eq0$ for all $x\in R$, we solve $x + 5=0$. So $x=-5$.

Step3: Determine intervals for increasing and decreasing

We consider the intervals separated by the critical point $x=-5$, which are $(-\infty,-5)$ and $(-5,\infty)$.
Take a test - point in $(-\infty,-5)$, say $x=-6$. Then $f^{\prime}(-6)=(-6 + 5)e^{12}=-e^{12}<0$, so $f(x)$ is decreasing on $(-\infty,-5)$.
Take a test - point in $(-5,\infty)$, say $x = 0$. Then $f^{\prime}(0)=(0 + 5)e^{0}=5>0$, so $f(x)$ is increasing on $(-5,\infty)$.

Step4: Find local extrema

Since $f(x)$ changes from decreasing to increasing at $x=-5$, by the first - derivative test, $f(x)$ has a local minimum at $x=-5$ and no local maximum.

Answer:

a. A. The critical point(s) of $f$ is/are $x=-5$.
b. $f$ is decreasing on $(-\infty,-5)$ and increasing on $(-5,\infty)$.
c. $f$ has a local minimum at $x=-5$ and no local maximum.