$$u(x)=\frac{m\omega^2x^2}{2}-\delta x-\frac{\alpha x^3}{3}$$ $\omega,\delta,\alpha$ are real and positive. Sketch the typical plots of $u(x)$ and identify extrema if any.
Now, I understand how to sketch a cubic polynomial. I take its derivative, find its discriminant (to check number of turning points), and then sketch accordingly.
However, the solution given for this problem is intriguing. They have done this:
We can define $D=m^2\omega^4-4\alpha\delta$ and $D'=m^2\omega^4-\dfrac{16}3\alpha\delta$. There are four possibilities:
I understand that $u'(x)=-\alpha x^2+m\omega^2x-\delta$, hence, $D_{u'(x)}=m^2\omega^4-4\alpha\delta$. But, I do not understand from where they got $D'$, and how they have used it to sketch the graphs.Please help me understand this. Thank you!
INPhO 2011 - Question paper (p21) and Solution (p3) - for reference
If you carefully observe, then you can see that $x=0$ is a solution for the cubic polynomial. So cancelling out $x$ term from the whole expression, we're left with a quadratic, solving that will help us find the remaining roots of the cubic equation. So $D'$ is nothing but the discriminant of quadratic remaining after cancelling out $x$.
The nature of $D'$ helps us in determining the nature of roots of cubic, directly. $D'<0$ implies no real root of the quadratic, I.e. only one root of cubic $x=0$. $D'=0$ suggests the remaining two roots are concurrent. Hence, the graph will touch the x-axis at that point, instead of intersecting. Similarly, you can understand other cases via graphs.