calculus
posted by michael .
Consider the function f(x)=4sqrtx+5 on the interval [4,5]. Find the average or mean slope of the function on this interval.
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.

1) 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