Question

Find the intervals where the function f(x)=xe^-x is concave up and concave down. Also find the point of inflection.

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{f(x)=x\,e^{-x}}$

$\underline{\textbf{To find:}}$

$\textsf{The intervals where f(x) concave up and concave down}$

$\textsf{and point of inflection}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{f(x)=x\,e^{-x}}$

$\mathsf{f'(x)=-xe^{-x}+e^{-x}}$

$\mathsf{f''(x)=-[-xe^{-x}+e^{-x}]-e^{-x}}$

$\mathsf{f''(x)=xe^{-x}-e^{-x}-e^{-x}}$

$\mathsf{f''(x)=xe^{-x}-2\,e^{-x}}$

$\mathsf{f''(x)=(x-2)e^{-x}}$

$\mathsf{f''(x)=0\implies\;(x-2)e^{-x}=0\implies\;x-2=0}$

$\implies\mathsf{x=2}$

$\textsf{Divide the real line into two intervals}\;\mathsf{(-\infty,2)\;\&\;(2,\infty)}$

$\begin{array}{|c|c|c|}\cline{1-3}\textsf{Interval}&\mathsf{Sign\;of\;f''(x)}&\mathsf{Concavity}\\\cline{1-3}(-\infty,2)&-&\mathsf{Concave\;downward}\\(2,\infty)&+&\mathsf{Concave\;upward}\\\cline{1-3}\end{array}$

$\textsf{The second derivative f''(x) changes it sign from - to + when}$

$\textsf{passing through x=2}$

$\therefore\textsf{Point of inflection is (2,f(2))}$

$\implies\mathsf{(2,2\,e^{-2})}$

$\implies\boxed{\mathsf{\left(2,\dfrac{2}{e^2}\right)}}$