determine the intervals on which the following function is concave up or concave down. identify any…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = e^{x}$ and $v=x - 3$. $u^\prime=e^{x}$ and $v^\prime = 1$. $f^\prime(x)=e^{x}(x - 3)+e^{x}\times1=e^{x}(x - 3 + 1)=e^{x}(x - 2)$

Step2: Find the second - derivative

Again use the product rule on $f^\prime(x)=e^{x}(x - 2)$. Let $u = e^{x}$ and $v=x - 2$, $u^\prime=e^{x}$ and $v^\prime = 1$. $f^{\prime\prime}(x)=e^{x}(x - 2)+e^{x}\times1=e^{x}(x - 2+1)=e^{x}(x - 1)$

Step3: Find the inflection point

Set $f^{\prime\prime}(x)=0$. Since $e^{x}\gt0$ for all real $x$, then $x - 1 = 0$, so $x = 1$.

Step4: Determine concavity intervals

Test intervals based on the inflection point $x = 1$. For $x\lt1$, let's choose $x = 0$. Then $f^{\prime\prime}(0)=e^{0}(0 - 1)=-1\lt0$, so the function is concave down on $(-\infty,1)$. For $x\gt1$, let's choose $x = 2$. Then $f^{\prime\prime}(2)=e^{2}(2 - 1)=e^{2}\gt0$, so the function is concave up on $(1,\infty)$.

Answer:

A. The function is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$