Consider the function f(x)=8sqrt(x)+7

Question

Consider the function f(x)=8sqrt(x)+7 on the interval [36] . 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 (36) such that f(c) is equal to this mean slope. For this problem, there is only one c that works. Find it. 

Answer 1

First we want the slope of the curve which is its derivative.

f(x) = 8*Sqrt(x) + 7

f'(x) = 4/Sqrt(x)

 

The Mean Value Theorem, says

f'(c) = (f(b) - f(a))/(b - a)  b = 6, a = 3

f'(c) = (8*Sqrt(6) + 7 -  (8*Sqrt(3) + 7))/(6 - 3)

= (8*Sqrt(6) + 7 - 8*Sqrt(3) - 7)/(3)

= (8*Sqrt(6) - 8*Sqrt(3))/(3)

f'(c)= (8/3)(Sqrt(6) - Sqrt(3))

 

f'(x) from a to b = f'(c)

4/Sqrt(x) = (8/3)(Sqrt(6) - Sqrt(3))

1/Sqrt(x) = (2/3)(Sqrt(6) - Sqrt(3))

1/x = (4/9)(Sqrt(6) - Sqrt(3))²

1/x = (4/9)(6 - 2(Sqrt(6)Sqrt(3)) +3)

1/x = (4/9)(9 - 2(Sqrt(6)Sqrt(3)))

1/x = 4 - (8/9)(Sqrt(6)Sqrt(3))

1/x = 4 - (8/3)(Sqrt(2))

x = 1/(4 - (8/3)(Sqrt(2)))

x = 4.3713 = c

Answer 2

I am assuming that the interval that you have in mind is [0,36].  

 

The mean value of the slope is  [ f(36) - f(0) ] / 36 =   48/36 = 4/3 .

 

The derivative of f is   4/ sqrt(x) .    To find c such that  f'(c) = 4/3,

 

set   4/sqrt(c) = 4/3.      The solution is c = 9   which is on the open interval (0,36)