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        riddles >> medium >> g(x) = g'(x)
        
(Message started by: ThudanBlunder on Jan 20^th, 2009, 6:02pm)

Title: g(x) = g'(x)
Post by ThudanBlunder on Jan 20^th, 2009, 6:02pm

If g(x) is a polynomial with a real root, show that there always exists an x such that g(x) = g'(x).

Title: Re: g(x) = g'(x)
Post by 1337b4k4 on Jan 21^st, 2009, 12:57am

[hide]wlog assume the leading coefficient of g(x) is positive.

1. if g(x) is an odd polynomial, then g(x) - g'(x) is also odd and has at least one real root

2. If g(x) is even and has a root tangent to the x axis, then that root satisfies g(x) = g'(x)

3. Otherwise, g(x) is even and must have at least two nonrepeating roots. Say the roots are, in increasing order, x1, x2, ... xk. we know g'(x1) is negative, g'(xn) is positive, thus g(x) - g'(x) is positive at x1, negative at xn, thus by IMT it must be 0 at some point.
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