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Math Help - Rolle's Theorem

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    Rolle's Theorem

    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.

    Any help? Thanks!
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    Quote Originally Posted by friday616 View Post
    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.

    Any help? Thanks!
    If we let f(x) = x^7+x^5+x^3+1, since f(-2) < 0 and f(0) > 0, and f is continuous, then by the intermediate value theorem, there's at least on root on [-2,0]. To prove there's only one root, suppose there are two, say f(a) = f(b) = 0. By Rolles theorem, there is a c in (a,b) where f'(c) = 0. Further, f'(x) must change sign crossing x = c (i.e. goes up levels out and comes down or vice-versa). Now 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.
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