Intermediate Value Theorem

**iLikeMaths** · April 18th 2009, 03:01 PM

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

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

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

if its the second equation then i know that the intervals are $[-2,-1]$ because $g(-1)>g(-2)$

**NonCommAlg** · April 18th 2009, 11:55 PM

Originally Posted by iLikeMaths

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

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

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

if its the second equation then i know that the intervals are $[-2,-1]$ because $g(-1)>g(-2)$

in general any polynomial $f(x)$ 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 $f,$ we have $\lim_{x\to{+ \infty}}f(x)=\pm \infty$ and $\lim_{x\to{-\infty}} f(x)=\mp \infty.$ so for $x \gg 0: \ f(x)f(-x) < 0.$ now apply IVT.

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