I need to examine and draw the graph of the function:
$${\displaystyle f\left(x\right)=\frac{x}{1+e^{\frac{1}{x}}}}.$$
I have problems with local minimum and maximum. First derivative is $$ f'\left(x\right)=\frac{e^{\frac{1}{x}}\left(1+x\right)+x}{x\left(1+e^{\frac{1}{x}}\right)^{2}}. $$
Now, I do not know how to solve $f'(x)>0$, $f'(x)<0$ and $f'(x)=0$. Any idea? What is the simplest way to do this?
A similar situation occurs with the second derivative of the function $$\log_{x^{2}-3x+2}2.$$
I found it is: $$f''(x)=\ln2\cdot\frac{\ln\left(x^{2}-3x+2\right)\left(2x^{2}-6x+5\right)+2\left(2x-3\right)^{2}}{\left(x^{2}-3x+2\right)^{2}\ln^{3}\left(x^{2}-3x+2\right)},$$
but how to solve $f''(x)>0$, $f''(x)<0$ and $f''(x)=0$. I suppose I don't have to know the exact values for x but how to sketch the graph without this.
As a hint:$$f''(x) =\begin{cases}\frac{\left(\mathrm{e}^\frac{1}{x}-1\right)\mathrm{e}^\frac{1}{x}}{x^3\cdot\left(\mathrm{e}^\frac{1}{x}+1\right)^3}=\frac{(e^{+}-1)e^+}{+}>0 & x > 0\\\text{undefined} & x=0\\\frac{\left(\mathrm{e}^\frac{1}{x}-1\right)\mathrm{e}^\frac{1}{x}}{x^3\cdot\left(\mathrm{e}^\frac{1}{x}+1\right)^3}=\frac{(e^{-}-1)e^-}{-}>0 & x > 0\end{cases} $$ so $f''>0$ and $f(x)$ is departed into two convex part for $0<x , x>0$