Thread: intermediate value theorem

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!

Factor out and then look at values of the resulting expression for very large x.

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

.

For very large positive x, the expression in the brackets is very close to 1, right? And the is a very large positive value.

For very large negative x, the expression in the brackets again is very close to 1 and the 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.