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September 20, 2014

September 20, 2014

Posted by **michael** on Thursday, November 12, 2009 at 1:31am.

By the Mean Value Theorem, we know there exists a c in the open interval (4,5) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.

- calculus -
**drwls**, Thursday, November 12, 2009 at 1:49am1) For the average value of the SLOPE of f(x) in that interval, calculate Integral of f'(x) from 4 to 5. You have to divide by the interval to get the average, but that is 1.

f'(x) = 2/sqrtx

Average value = INTEGRAL 2 x^-1/2 dx

from 4 to 5

= 4 (sqrt 5 - sqrt 4) = 0.9443

2) Now that you have the average value of f'(x), find the value of c such that f'(c) = f'(x)(average)

2/sqrtc = 0.9443

sqrt c = 2.118

c = 4.486

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