Find c for Average Value of Graph to be 0: Why c=4?

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Homework Help Overview

The discussion revolves around determining the value of c>0 for which the average value of a graph on the interval [-5, c] equals zero. Participants are exploring the implications of the Mean Value Theorem in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster suggests that c should be 4, reasoning that the areas under the graph cancel out at this point. Other participants mention the Mean Value Theorem and its relevance to the problem, questioning its application and implications.

Discussion Status

Participants are actively engaging with the problem, with some suggesting the use of the Mean Value Theorem for integrals. There is a recognition of the need to clarify the relationship between the average value of the function and the integral over the specified interval.

Contextual Notes

There is a pending approval for the graph that is central to the discussion, which may affect the clarity of the problem setup. Some participants express uncertainty about the Mean Value Theorem and its application to this scenario.

chukie
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For what value of c>0 is the average value of the graph on [-5,c] equal to zero and why?

Graph is attached.


I thought that that c should be 4 since at that point the areas cancel out. Not sure though. Could someone help me? Thanks!
 

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I don't see a graph since it's still pending approval, but use the Mean Value Theorem(for integrals) to solve for c.
 
ive managed to get the graph
 
i also looked at the mean value theorem. it deals with f'(x)=0 ?
 
Ah, ok. Yes, c=4 since f is a continuous function on [-5,c], then the average value of f from -5 to c = [tex]\frac{1}{c-(-5)}\[ \int_{-5}^{c} f(x)\,dx.\][/tex]. Since the average value = 0, then it's just the integral from -5 to 4(since it's a graph, yeah the areas cancel out)

i also looked at the mean value theorem. it deals with f'(x)=0 ?
No, check out the MVT for integrals also known as the average value of a function.
 
Last edited:
okay great! thanks so much =)
 

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