Mean value theorm, what did i do wrong?

[apiwowar]

so the function is f(x) = 8sqrt(x) + 8

part a says consider the function on the interval [3,5] and find the average slope on this interval

i know how to solve this using the mean value theorem f'(c) = f(b)-f(a)/(b-a)

and you get 4sqrt(5) - 4sqrt(3)

part be then asks, "By the Mean Value Theorem, we know there exists a c in the open interval (3,5) such that f'(c) is equal to the mean slope. for this problem there is only one c that works. find it.

so what i did was first find f'(x) which is 4/sqrt(x) then i set that equal to the average slope that i figured out in the first part so you get:

4/sqrtc = 4sqrt5 - 4sqrt3

to solve for c then i divided both sides by 4 to get:

1/sqrtc = sqrt5 - sqrt3

then i squared both sides and got c = 1/2 but that can't be right since c has to be between (3,5). doiing it another way i got c = 2 but that again can't be right

hopefully someone can spot out a mistake that i made, i would really appreciate it.

thanks

 

[apiwowar]

nevermind, i figured out what i was doing wrong, i didnt square (sqrt5-sqrt3) the right way