summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of…

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of f(x)=2(4 - x)e^x. find the location of any local extrema of f(x). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. there is a local minimum at x = . there is no local maximum. (type an integer or a decimal. use a comma to separate answers as needed.) b. there is a local maximum at x = . there is no local minimum. (type an integer or a decimal. use a comma to separate answers as needed.) c. there is a local maximum at x = and there is a local minimum at x = . (type an integer or a decimal. use a comma to separate answers as needed.) d. there are no local extrema.

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of f(x)=2(4 - x)e^x. find the location of any local extrema of f(x). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. there is a local minimum at x = . there is no local maximum. (type an integer or a decimal. use a comma to separate answers as needed.) b. there is a local maximum at x = . there is no local minimum. (type an integer or a decimal. use a comma to separate answers as needed.) c. there is a local maximum at x = and there is a local minimum at x = . (type an integer or a decimal. use a comma to separate answers as needed.) d. there are no local extrema.

Answer

Explanation:

Step1: Find the derivative of $f(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 2(4 - x)$ and $v = e^{x}$. $u^\prime=-2$ and $v^\prime = e^{x}$. Then $f^\prime(x)=-2e^{x}+2(4 - x)e^{x}=2e^{x}(3 - x)$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. Since $e^{x}>0$ for all real $x$, we solve $2e^{x}(3 - x)=0$. $3 - x = 0$, so $x = 3$.

Step3: Determine the nature of the critical point

Use the first - derivative test. Choose a test point to the left of $x = 3$, say $x = 2$. Then $f^\prime(2)=2e^{2}(3 - 2)=2e^{2}>0$. Choose a test point to the right of $x = 3$, say $x = 4$. Then $f^\prime(4)=2e^{4}(3 - 4)=-2e^{4}<0$. Since $f^\prime(x)$ changes sign from positive to negative at $x = 3$, $f(x)$ has a local maximum at $x = 3$. There is no local minimum.

Answer:

B. There is a local maximum at $x = 3$. There is no local minimum.