Show that $x^4+2x^2-6x+2=0$ when $x\in\mathbb{R}$ has exactly two solutions.
I first showed that the IVT guarantees that there exists at least one zero in $(0,1)$ and at least one zero in $(1,2)$. I then was going to apply Rolle's theorem twice to show uniqueness in each interval.
My goal was to assume contradiction in each interval but I was hoping that $f'(x) > 0$, but it is not.
Where did I go wrong, and how can I fix it?
Best Answer
After your findings so far, it suffices to show that the derivative $$ f'(x)=4x^3+4x-6$$ has only one real root. For this again, it is sufficient to observe that the second derivative $$ f''(x)=12x^2+4$$ is strictly positive.
In other words, your Rolle aproach should work: If there were three real roots $x_1<x_2<x_3$ of $f$, we'd have roots $\xi_1,\xi_2$ of $f'$ with $\xi_1<x_2<\xi_2$ and then a root $\eta_1$ of $f''$.