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 $v = x - 10$. $u^\prime=-e^{x}$, $v^\prime = 1$. $f^\prime(x)=-e^{x}(x - 10)-e^{x}=-e^{x}(x - 10 + 1)=-e^{x}(x - 9)$.

Step2: Find the second - derivative

Again use the product rule. Let $u=-e^{x}$ and $v=x - 9$. $u^\prime=-e^{x}$, $v^\prime = 1$. $f^{\prime\prime}(x)=-e^{x}(x - 9)-e^{x}=-e^{x}(x - 9+1)=-e^{x}(x - 8)$.

Step3: Find where $f^{\prime\prime}(x)>0$ (concave up)

Since $e^{x}>0$ for all $x\in R$, we solve $-e^{x}(x - 8)>0$. Divide both sides by $-e^{x}$ (and reverse the inequality sign because $-e^{x}<0$). We get $x - 8<0$, so $x<8$. The function is concave up on $(-\infty,8)$.

Step4: Find where $f^{\prime\prime}(x)<0$ (concave down)

Solve $-e^{x}(x - 8)<0$. Divide both sides by $-e^{x}$ (reverse the inequality sign). We get $x - 8>0$, so $x>8$. The function is concave down on $(8,\infty)$.

Step5: Find inflection points

Set $f^{\prime\prime}(x) = 0$. Since $e^{x}\neq0$ for all $x\in R$, we set $x - 8=0$. So $x = 8$ is the inflection point.

Answer:

The function is concave up on $(-\infty,8)$ and concave down on $(8,\infty)$. An inflection point occurs at $x = 8$. A. The function is concave up on $(-\infty,8)$ and concave down on $(8,\infty)$. A. An inflection point occurs at $x = 8$.