Function Inequality: Show t Exists for f(x) & x^2

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Homework Help Overview

The problem involves a continuous function f(x) defined from the reals to the reals, with a limit condition that relates f(x) to x^2 as x approaches positive and negative infinity. The goal is to demonstrate the existence of a value t such that a specific inequality involving f(x) and x^2 holds for all real numbers x.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the limit condition on the behavior of f(x) relative to x^2. Some suggest that showing the function x^2 + f(x) has a minimum could be a viable approach. Others question the concept of boundedness and its relevance to the problem.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of boundedness and its implications for the function. Some have offered insights into the properties of continuous functions, while others are questioning specific examples and their relevance to the inequality to be proven.

Contextual Notes

There is a mention of a specific function, f(x) = x, which raises concerns about the existence of a minimum for the expression in question. Participants are grappling with the definitions and implications of boundedness in the context of the problem.

Kruger
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Homework Statement



Given a continuous function f(x):R->R with

lim(f(x)/x^2)=0, x-->+-infinity

Show that then an element t exist such that:

x^2+f(x)>=t^2+f(t) for every x in R.

Homework Equations



-> The mathematical definition of continuous and limes
(but I really don't know if these are needed)

The Attempt at a Solution



I really thought hours on that problem but didn't find a solution. I've no really good attempt. Well, I know that f(x) has to increase less rapidly than x^2 accordint to lim(f(x)/x^2)=0, x-->+-infinity.
 
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You just need to show that the function x^2+f(x) has a minimum. Since f(x) is continuous, it is bounded on any finite interval, and the first condition shows f(x) goes to zero as x->infinity, so you can show that f(x) is bounded for all x. Combine this with the fact that x^2 is bounded below.
 
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In theory f(x) could be a function like that one in my picture. I don't really understand what you mean with "bounded". I mean in what case are they bounded?
 
"Bounded" just means that it has upper and lower bounds: that there exist numbers m and M such that m<= f(x)<= M for all x. If lim f(x)/x^2= 0 for x-> +infinity, then There exist X such that if x> X, |f(x)/x^2|< 1: i.e. -1< f(x)/x^2. Similarly, since lim f(x)/x^2= 0 for x-> -infinity there exist X' so that the same is true for x< X'. Let m be the smallest value of f(x)/x^2 for X'<= X<= X and you know f(x)/x^2<= m (if m> -1, take m= -1) for all x.
 
But consider f(x)=x. Then f(x)+x^2 has no real minimum and lim f(x)/x^2=x/x^2=1/x=0. So I must show that |f(x)+x^2| has a minimum.

By the way:
HallsofIvy: I understand your explanation but how would you integrate this in a proove of x^2+f(x)>=t^2+f(t) for every x in R?
 
Kruger said:
But consider f(x)=x. Then f(x)+x^2 has no real minimum and lim f(x)/x^2=x/x^2=1/x=0. So I must show that |f(x)+x^2| has a minimum.

f(x)+x^2=x^2+x, which has a minimum at x=-1/2.
 
uffff, you're right, man, what's up with me :) . If I tell that x^2+x has no minimum, how can I ever solve this :) (seems I'm disturbed).
 

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