1. ## 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 $\displaystyle g(-1)>g(-2)$

2. Originally Posted by iLikeMaths
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.