If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b)such that  f '(c) = f(b)-f(a)/b-a.   1. f(x)=square root of 1-x, [-8,1]   I understand that The mean...

If the Mean Value Theorem can be applied, find all values of c in the open interval

(a, b)such that  f '(c) = f(b)-f(a)/b-a.
 
1. f(x)=square root of 1-x, [-8,1]
 
I understand that The mean value theorem can be applied, but how do you find what C is equal too? The square root is confusing me and I cannot find an example to guide me through this question. I am not seeking an answer to this, because i prefer solving things on my own, so can anyone provide an explanation to an example similar to this problem? It would be greatly appreciated.

Asked on by jcamacho55

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nick-teal | High School Teacher | (Level 3) Adjunct Educator

Posted on

The mean value theorem tells you that a certain slope is possible over an interval.

What this question is asking you is to find all the points (values of c) where this slope exists over the interval.  There can be one or more points with the mean value theorem slope.

I will do another problem.

`f(x) = x^2` on the interval [-2, 1]

I find the slope between those two points (-2, 4) and (1,1)

`(4-1)/(-2-1) = -1` 

I know there will be at least one point with slope -1.

`f'(x) = 2x`

I want to find all values of x where f'(x) is -1

`-1 = 2x`

x = -1/2

Therefore there is one value of c, and it is -1/2

Hope this helps you solve your problem!

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