1. ## Mean Value Thereom

Hi, I want to show that this function satisfies the Mean Value Thereom however, I am stuck when I try to obtain c. The function is f (x) = x/(x+2) on the interval [1,4].

So far I have:

That because it is a rational fuction, f is continuous on [a,b] and differentiable on (a,b).

f(4) - f(1) = f'(c)(4-1)
1/3 = f ' (c)*3
f '(c) = 1/9

f'(x) = 2/[(x+2)^2]

however when i try to set f'(x) = 1/9, i do not obtain a value within [1,4]

Thank you so much for your help!

2. Originally Posted by aquaglass88
f(4) - f(1) = f'(c)(4-1)
1/3 = f ' (c)*3 f '(c) = 1/9
f'(x) = 2/[(x+2)^2]
however when i try to set f'(x) = 1/9, i do not obtain a value within [1,4]
I found a solution in [1,4].
Square the term and solve. You should find the x is about 2.2

3. you mean square the equation 2/[(x+2)^2]?

4. $\frac{2}{{x^2 + 4x + 4}} = \frac{1}{9}$

5. but i get negative 2.2, and that isn't in [1,4]
[4-sq(72)]/2

6. oh wait nm it's -4 thats why i messed up sorry!

thanks so much!