This function has a null point, but I can't compute it from equation $f(x)=0$ which gives $$\frac{x+e^x}{x-e^x}=0$$ $$x+e^x=0$$ How to compute this equation?
Extreme points can be computed from equation $f'(x)=0$, $$\frac{(2x-2)e^x}{e^{2x}-2xe^x+x^2}=0$$ which gives $$L_{min}(1,-2.16)$$ is the only extreme point of $f(x)$.
Inflection points can't be computed from equation $f''(x)=0$ which gives $$-\frac{(2x-4)e^{2x}+(2x^2-4x+4)e^x}{e^{3x}-3xe^{2x}+3x^2e^{2x}-x^3}=0$$ Could someone show how to find inflection points of $f(x)$?
Thanks for replies.