# Using both intermediate value theorem and rolle's theorem

For example, to prove that the function $$x^8+x−1=0$$ has exactly two roots, we first prove that $$f$$ has at most 2 real roots by using differentiation. $$f'(x)=8x^7+1=0$$, by using that the function has two roots and therefore it should have at least one point such that $$f′(c)=0$$ and then we use the IVT to prove that it has exactly two roots.

However, the derivative of equation $$x^4−6x^2−8x+1=0$$ has 2 roots but still, I should try to prove that it has exactly two roots. If its derivative had one solution, then I could first prove that it has two roots at most as I did but it has two roots, so how should we approach the problem?

The derivative $$4x^3-12x-8$$ factors into $$(x+1)^2(x-2)$$. Thus $$-1$$ is a double root at which the derivative doesn't change sign. Thus $$x^4-6x^2-8x+1$$ is monotonically decreasing for $$x\lt2$$ and increasing for $$x\gt2$$. It's negative at $$x=2$$ and goes to $$\infty$$ for $$x\to\pm\infty$$, so it has exactly two roots.