Prove that the equation x^7+x^5+x^3+1=0 as exactly one real solution. You should use Rolle's Theorem at some point in the proof.
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If we let $\displaystyle f(x) = x^7+x^5+x^3+1$, since $\displaystyle f(-2) < 0$ and $\displaystyle f(0) > 0,$ and $\displaystyle f$ is continuous, then by the intermediate value theorem, there's at least on root on $\displaystyle [-2,0]$. To prove there's only one root, suppose there are two, say $\displaystyle f(a) = f(b) = 0$. By Rolles theorem, there is a $\displaystyle c$ in $\displaystyle (a,b) $ where $\displaystyle f'(c) = 0.$ Further, $\displaystyle f'(x) $ must change sign crossing $\displaystyle x = c$ (i.e. goes up levels out and comes down or vice-versa). Now $\displaystyle f'(x) = 7x^6+5x^4+3x^2 \ge 0$ (it never changes sign). Thus, we have a contradiction - there can't be two roots.