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 $v=x - 3$. $u^\prime=e^{x}$, $v^\prime = 1$. Then $f^\prime(x)=e^{x}(x - 3)+e^{x}=e^{x}(x - 3+1)=e^{x}(x - 2)$.

Step2: Find the second - derivative

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

Step3: Determine concavity

Set $f^{\prime\prime}(x)=0$. Since $e^{x}>0$ for all $x\in R$, then $x - 1=0$, so $x = 1$. When $x<1$, $f^{\prime\prime}(x)=e^{x}(x - 1)<0$, so the function is concave down on $(-\infty,1)$. When $x>1$, $f^{\prime\prime}(x)=e^{x}(x - 1)>0$, so the function is concave up on $(1,\infty)$.

Step4: Find inflection points

An inflection point occurs where the concavity changes. Since $f^{\prime\prime}(x)$ changes sign at $x = 1$, and $f(1)=e^{1}(1 - 3)=-2e$. The inflection point is at $x = 1$.

Answer:

A. The function is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$. A. An inflection point occurs at $x = 1$.