Prove Existence: F(x) = (x-a)^2(x-b)^2 + x

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Homework Help Overview

The problem involves the function F(x) = (x-a)^2(x-b)^2 + x, with the goal of demonstrating that the output (a+b)/2 exists for some value of x. The discussion centers around the properties of this function and its behavior at specific points.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the function's zeros and its output, with some questioning how to effectively demonstrate the existence of a specific output. There is also discussion about the implications of setting a equal to b and the informal nature of "showing" versus proving.

Discussion Status

Participants have offered hints regarding the function's values at specific points, such as F(a) and F(b), and have suggested the use of the Intermediate Value Theorem as a potential approach. Multiple interpretations of how to approach the problem are being explored.

Contextual Notes

There is a mention of the function's behavior when a equals b, which raises questions about the assumptions underlying the problem. The nature of the term "show" in the context of the problem is also under discussion.

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Homework Statement


[tex]Let F(x) = (x-a)^2(x-b)^2 + x[/tex]. Show that the output [tex]\frac{a+b}{2}[/tex] exists for some value x.

Homework Equations


Quadratic formula. [tex]x^2 \geq 0[/tex].

The Attempt at a Solution


Hmm I've tried setting the two equal but that doesn't look nice (if I multiply everything out). It's easy to find the zeros of F(x) so there might be someway to relate to that? If someone could just give me a hint at a good first step for showing the existence of a certain output of a function.
 
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Actually, I just edited it since there was an x in there. Now if a = b, then the output (a+b)/2 has to exist right? I'm not sure how to "show" it though. Show is just a bit more informal than a proof right?
 
F(a)=a and F(b)=b. That's a pretty good hint.
 
So invoke the Intermediate Value Theorem?
 
snipez90 said:
So invoke the Intermediate Value Theorem?

Exactly.
 

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