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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?

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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.

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