intermediate value theorem

**nngktr** · May 31st 2010, 11:17 AM

Let f (x) be a monic polynomial of odd degree. Prove that for some A < 0,
f (A) < −1 and for some B > 0, f (B) > 1. Deduce that every polynomial of odd degree has a real root.

Suppose f(x) =x^(2n+1)+a2n x^(2n)...+a1x+a0, but I've no clue how to go from here. Could anyone please give me some hints? Any help is appreciated!

**JG89** · May 31st 2010, 11:31 AM

Factor out $x^{2n + 1}$ and then look at values of the resulting expression for very large x.

**nngktr** · May 31st 2010, 11:51 AM

f(x)=(x-1)(x^2n+x^(2n-1)...+1)+1+...>1 when x is sufficiently large
f(x)=(x+1)(x^2n-x^(2n-1)+...+1)-1+...<-1 when x is sufficiently small
Is that what you are talking about? I'm still a bit unsure

**JG89** · May 31st 2010, 10:00 PM

$f(x) = x^{2n + 1} + a_{2n} x^{2n} + \cdots + a_1x + a_0 = x^{2n+1} (1 + a_{2n} \frac{1}{x} + a_{2n - 1} \frac{1}{x^2} + \cdots + a_1 \frac{1}{x^{2n}} + a_0 \frac{1}{x^{2n+1}})$ .

For very large positive x, the expression in the brackets is very close to 1, right? And the $x^{2n+1}$ is a very large positive value.

For very large negative x, the expression in the brackets again is very close to 1 and the $x^{2n+1}$ is a very large negative value, since the exponent is negative.

So you know that for large enough positive x, f(x) > 0 and for large enough negative x, f(x) < 0. Now apply the intermediate value theorem.

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