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

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 6x$ and $v=e^{- 0.5x}$. $u^\prime=6$ and $v^\prime=-0.5e^{-0.5x}$. So $f^\prime(x)=6e^{-0.5x}+6x(-0.5e^{-0.5x})=6e^{-0.5x}(1 - 0.5x)$.

Step2: Find the second - derivative

Use the product rule again on $f^\prime(x)=6e^{-0.5x}(1 - 0.5x)$. Let $u = 6(1 - 0.5x)$ and $v = e^{-0.5x}$. $u^\prime=-3$ and $v^\prime=-0.5e^{-0.5x}$. $f^{\prime\prime}(x)=-3e^{-0.5x}+6(1 - 0.5x)(-0.5e^{-0.5x})=e^{-0.5x}(-3 - 3 + 1.5x)=e^{-0.5x}(1.5x - 6)$.

Step3: Find the inflection points

Set $f^{\prime\prime}(x)=0$. Since $e^{-0.5x}\gt0$ for all real $x$, we solve $1.5x - 6 = 0$. $1.5x=6$, so $x = 4$.

Step4: Determine concavity

Test intervals. Consider the intervals $(-\infty,4)$ and $(4,\infty)$. For $x\lt4$, say $x = 0$, then $f^{\prime\prime}(0)=e^{0}(0 - 6)=-6\lt0$. So $f(x)$ is concave downward on $(-\infty,4)$. For $x\gt4$, say $x = 5$, then $f^{\prime\prime}(5)=e^{-2.5}(7.5 - 6)=1.5e^{-2.5}\gt0$. So $f(x)$ is concave upward on $(4,\infty)$.

Answer:

B. The function is concave upward on $(4,\infty)$. It is concave downward on $(-\infty,4)$.