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Thread: Intermediate Value Theorem

  1. #1
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    Intermediate Value Theorem

    im supposed to use the IVT to prove that the equation
    $\displaystyle g(x)= 1+x + \frac{x^2}{2}+....+\frac{x^{n-1}}{n-1}+\frac{x^n}{n}$ has at least one real solution. where $\displaystyle n$ is and odd positive integer and $\displaystyle g(x)=0$.

    im not sure if when $\displaystyle n=3$ the equation looks like $\displaystyle g(x)=1+x+\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^3}{3}$

    or
    $\displaystyle g(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3}$

    if its the second equation then i know that the intervals are $\displaystyle [-2,-1]$ because $\displaystyle g(-1)>g(-2)$
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  2. #2
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    Quote Originally Posted by iLikeMaths View Post
    im supposed to use the IVT to prove that the equation
    $\displaystyle g(x)= 1+x + \frac{x^2}{2}+....+\frac{x^{n-1}}{n-1}+\frac{x^n}{n}$ has at least one real solution. where $\displaystyle n$ is and odd positive integer and $\displaystyle g(x)=0$.

    im not sure if when $\displaystyle n=3$ the equation looks like $\displaystyle g(x)=1+x+\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^3}{3}$

    or
    $\displaystyle g(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3}$

    if its the second equation then i know that the intervals are $\displaystyle [-2,-1]$ because $\displaystyle g(-1)>g(-2)$
    in general any polynomial $\displaystyle f(x)$ of odd degree and with real coefficients has a real root. you want to prove it using IVT:

    see that, depending on the sign of the leading coefficient of $\displaystyle f,$ we have $\displaystyle \lim_{x\to{+ \infty}}f(x)=\pm \infty$ and $\displaystyle \lim_{x\to{-\infty}} f(x)=\mp \infty.$ so for $\displaystyle x \gg 0: \ f(x)f(-x) < 0.$ now apply IVT.
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