Math Is Fun Forum

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

real analysis help

I have two questions, help with either would be greatly appreciated.

1. f is continuous on [a, infinity) and lim f(x) exists thn f is bouned on [a, infinity)
2. Let f(x) be any polynomial of degree at least 2, all of whose roots are real and distinct. Prove that all of the roots of f'(x) must be real. What happens if some roots of f are multiple roots?

For #2 I'm certain that I will have to use either Rolles Theorem of Fermat's Theorem on Extrema. Ideas?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: real analysis help

#2
If f has degree n, let the roots be

in increasing order. Consider the interval

. By Rolle’s theorem,

such that

. This gives n−1 distinct roots

for the derivative f′. Indeed, these are all the roots of f′, since f′ has degree n−1 and so cannot have more than n−1 roots. QED.

If a root of f is repeated, then that root will also be a root of f′. Indeed, if the root is repeated k times for f, it will be repeated k−1 times for f′. In any case, all the roots of f′ are still real.

Last edited by JaneFairfax (2008-02-24 11:56:38)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: real analysis help

#1
We’ll prove the contrapositive: if f is not bounded, then f does not tend to a real limit as x tends to infinity.

Consider the sequence of closed intervals

A continuous function must be bounded on a closed interval. Also, it must attain a maximum and minimum in that interval, i.e.

such that

If the sequence

is bounded above, then f would be bounded above. Likewise if the sequence

is bounded below, then f would be bounded below. Since we are assuming that f is not bounded, either the

’s are not bounded above or the

’s are not bounded below (possibly both). Hence

does not exist.