function with two horizontal asymptotes and f'(0) = 0

Solution 1:

Another way to get two different horizontal asymptotes is to have a square root: $$y=\frac{x}{\sqrt{x^2+1}}$$

To get this to have a derivative of $0$ at $x=0$, you need to just increase the exponents to higher odd powers (even powers will result in one horizontal asymptote): $$y=\frac{x^3}{\sqrt{x^6+1}}$$

Then if you need to have the $x=0$ point to actually be a maximum, add any function that tends to $0$ at both plus and minus infinity that also has a derivative of $0$ at $x=0$, e.g. $e^{-x^2}$, with a sufficiently large coefficient: $$y=\frac{x^3}{\sqrt{x^6+1}}+5e^{-x^2}$$

Solution 2:

$$\arctan(x)+\frac{2}{x^2+2x+2}$$

The idea is: $\arctan(x)$ has two different imits at $-\infty$ and $+\infty$. But it does not have derivative $0$ at $0$. And $\frac{1}{(x+1)^2+1}$ has limits $0$ at $-\infty$ and $+\infty$ and nonzero derivative at $0$. So find a linear combintation of these with derivative $0$ at $0$.

Solution 3:

This works: $$ f(x) = 5 e^{-x^2}+2 \tanh \left(\frac{x}{2}\right)-\tanh (x). $$ The graph: