for the function: f(x) = xe^x find the inflection point. (?, )

Explanation:

Step1: Find the first - derivative

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

Step2: Find the second - derivative

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

Step3: Find the x - coordinate of the inflection point

Set $f^{\prime\prime}(x)=0$. Since $e^{x}\gt0$ for all real $x$, then $(x + 2)e^{x}=0$ implies $x+2 = 0$, so $x=-2$.

Step4: Find the y - coordinate of the inflection point

Substitute $x = - 2$ into the original function $f(x)=xe^{x}$. Then $f(-2)=-2e^{-2}=-\frac{2}{e^{2}}$.

Answer:

$(-2,-\frac{2}{e^{2}})$