Use linear stability analysis to classify the fixed points of the following system. $f(x)=ax-x^3 $ where a can be positive, zero or negative.
I have found that for $a>0$ we have $2$ fixed points For $a=0$ only $x=0$ is a fixed point at which it is stable
I don't understand how the solution for $a<0. $
I am not sure why $x=0$ is only the solution
any help will be appreciated thank u!