How Does the Mean Value Theorem Apply When a Function's Derivative Is Zero?

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Homework Help Overview

The discussion revolves around the application of the Mean Value Theorem (MVT) in the context of a function whose derivative is zero. Participants are exploring the implications of this scenario and how it relates to the theorem's statement.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the implications of having a derivative of zero and how it relates to the MVT. Questions arise about the correct application of the theorem and the interpretation of the variables involved, particularly regarding the role of 'c' and the function's values at points 'a' and 'b'.

Discussion Status

The discussion is active, with participants questioning the setup and the assumptions related to the derivative being zero. Some guidance has been offered regarding the implications of the derivative being zero, but there is no explicit consensus on how to proceed with the problem.

Contextual Notes

There is uncertainty regarding the specific function and the values of 'a' and 'b', which are not provided in the discussion. Participants are navigating these constraints while trying to apply the MVT.

ACLerok
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The Mean Value Theroem there is a number c in the interval (a,b) such that
f'(c)=f(b)-f(a)/b-a or f(b)-f(a)=f'(c)(b-a)

Okay, say a function has a derivative of zero. I'm supposed to explain what the Mean Value Theroem states about a function whose derivative is zero.

Any takers?
 
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f'(c)=f(b)-f(a)/b-a

This isn't right; you forgot some parentheses.


Anyways, you're told that the derivatve is zero, right? Well, plug in zero for the derivative and what do you get?
 
i plug in 0 for c it would be 0. so what?
 
If the derivative of a function is zero at all points, then the total change in value of the function is going to be zero. So now, does it really matter which point you pick? It's going to be zero either way!

cookiemonster
 
You were told the derivative of f is zero, right? Is c the derivative of f? (No) So why plug 0 in for c? Plug 0 in for the derivative of f.
 
Last edited:
Hurkyl said:
You were told the derivative of f is zero, right? Is c the derivative of f? (No) So why plug 0 in for c? Plug 0 in for the derivative of f.
plug 0 in where?
 
f'(c) = 0

so we get

0 = (f(b) - f(a))/(b-a)

Now solve for c.

cookiemonster
 
how do i know what the functino is or a and b? do i just solve for it in terms of f a and b?
 

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