locate the critical points of the following function. then use the second derivative test to determine…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = e^{x}$ and $v=x + 6$. $u^\prime=e^{x}$, $v^\prime = 1$. Then $f^\prime(x)=e^{x}(x + 6)+e^{x}=e^{x}(x + 7)$. Set $f^\prime(x)=0$, since $e^{x}\gt0$ for all $x\in R$, then $x+7 = 0$, so $x=-7$.

Step2: Find the second - derivative

Differentiate $f^\prime(x)=e^{x}(x + 7)$ using the product rule again. Let $u = e^{x}$ and $v=x + 7$, $u^\prime=e^{x}$, $v^\prime = 1$. Then $f^{\prime\prime}(x)=e^{x}(x + 7)+e^{x}=e^{x}(x + 8)$.

Step3: Apply the second - derivative test

Evaluate $f^{\prime\prime}(x)$ at the critical point $x=-7$. $f^{\prime\prime}(-7)=e^{-7}(-7 + 8)=e^{-7}\gt0$.

Answer:

The critical point is $x=-7$. Since $f^{\prime\prime}(-7)\gt0$, the function $f(x)$ has a local minimum at $x=-7$.