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Topic: g(x) = g'(x) (Read 479 times) |
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ThudnBlunder
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Uberpuzzler
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g(x) = g'(x)
« on: Jan 20th, 2009, 6:02pm » |
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If g(x) is a polynomial with a real root, show that there always exists an x such that g(x) = g'(x).
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teekyman
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Re: g(x) = g'(x)
« Reply #1 on: Jan 21st, 2009, 12:57am » |
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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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| « Last Edit: Jan 21st, 2009, 1:00am by teekyman » |
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