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

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The Mean Value Theorem states that for a function f, there exists a number c in the interval (a, b) such that f'(c) equals the average rate of change of the function over that interval. When the derivative of a function is zero, it indicates that the function does not change value over the interval, leading to the conclusion that f(b) - f(a) equals zero. Therefore, regardless of the specific values of a and b, the total change in the function's value remains zero. The discussion emphasizes the importance of correctly interpreting the derivative and its implications for the function's behavior. Understanding these concepts is crucial for applying the Mean Value Theorem effectively.
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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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