equation: $x' = x^3 +ax - b$
the task is to show that in plane (a,b) there is curve C which divide the plane into 2 parts A and B where values from A will give 1 Fixed point for the given equation and in B there will be 3 Fixed points for the given equation.
now im having some problem in investigating this problem and i was wondering if any of you could give me some hint on how to do so.
what i've done so far:
Fixed point means $x' = 0$ so we have the next equation
$x^3 = -ax + b$ so we can see geometrically that for $a>0$
we have only 1 intersection point which means only 1 fixed point for the equation
so we covered half of the (a,b) plane by now ($a > 0$).
now when $a < 0 $ there could be couple of cases which im having hard time to find.
any idea would be much appreciated.
Edit: Thanks to MrYouMath, for any one who wonders you can see the intersections between 2 points right here
Hint: $x'=0=x^3+ax-b$ Is a cubic equation, which can have 1 or 3 real solutions. You can even calculate them with Cardano's formula.
In order to show that there is always at least one real solution look at $p(x)=x^3+ax-b$ for $|x|\to \infty$ and use the intermediate value theorem.
If $p(x)$ has at least one real solution $x_1$ then we can rewrite the polynomial as $p(x)=(x-x_1)(x^2+\alpha x+ \beta)$. The quadratic part has again real coefficients and this can have no real solution, or two real solution.
Edit: If you want to find the curve you can use the cubic discriminant $\Delta = -4a^3-27b^2$. If the cubic discriminant is zero then we have a double zero, which is the case that is separating the one + two real solutions case to the one real + 2 complex conjugate solutions. So the curve of separation is given by the implicit curve $F(a,b)=4a^3+27b^2=0$.