Question

Calculus: Verify that the function f(x) = 1/(x+1) satisfies the Mean Value Theorem on interval [0,2]. Then find all values of "c" such that

f(2) - f(0)
--------
2 - 0 = f ' (c)

Answer 1

\[
f'(x) = - \frac 1{(x+1)^2}
\]

Answer 2

My work so far:
- f(x) is rational, and is continuous on {x|x =/= -1}. Then f(x) is continuous on [0,2].
- f' (x) = -1/(x+1)^2. f '(x) exists on {x|x =/= -1}. (MVT is fulfilled).
- -1 / (c + 1)^2 = ( f(2) - f(0) ) / 2

Answer 3

\[
\frac{1}{2} (f(2)-f(0))=-\frac{1}{3} = -\frac 1 {(1+c)^2
}
\]

Answer 4

wait. I think I know where I made my mistake
I use the original function for f(2) and f(0)

Answer 5

Solving the above gives you
\[
c= -1 + \sqrt 3\\
c=-1-\sqrt 3
\]
you take the positive root.