Answer:
Analyzed & Sketched
Step-by-step explanation:
We are given the function: [tex]f(x) = \frac{1}{xe^x}[/tex]
So, we will find first derivative for finding the interval of increase and decrease.
[tex]f'(x)=\frac{-e^x-xe^x}{x^2e^{2x}} =-\frac{e^{-x}(x+1)}{x^2}[/tex]
it is increasing on (-∞, -1) and decreasing on (1,0) ∪ (0, ∞).
For concavity, we need to find the second derivative.
[tex]f''(x)=\frac{(-e^x-e^x-xe^x)x^2e^{2x}+(e^x-xe^x)(2xe^{2x}+x^2e^{2x})}{x^4e^{4x}} =\frac{e^{-x}\left(x^2+2\left(x+1\right)\right)}{x^3}[/tex]
it is concave down on (-∞, 0) and concave up on (0, ∞).
x = 0 is vertical asymptote.
Sketch is given in the attachment.
