# Find Point c that satisfies the Mean Value Theorem

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1. Apr 6, 2017

### Schaus

1. The problem statement, all variables and given/known data
Find the point "c" that satisfies the Mean Value Theorem For Derivatives for the function
$f(x) = \frac {x-1} {x+1}$ on the interval [4,5].
2. Relevant equations
$x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}$
$f'(c) = \frac { f(b) - f(a)} {b-a}$

3. The attempt at a solution
I found the derivative
$f'(x) = \frac {2} {(x+1)^2}$
a) $f(4) = \frac {4-1} {4+1} = \frac {3} {5}$
b) $f(5) = \frac {5-1} {5+1} = \frac {2} {3}$
Substituting in my values
$\frac {2} {(x+1)^2} = \frac { \frac {2} {3} - \frac {3} {5}} {5-4}$
$\frac {2} {(x+1)^2} = \frac {1} {15}$
$30 = c^2 +2c +1$
$0 = c^2 +2c -29$
$x = \frac {-2 \pm \sqrt{2^2 -(4)(1)(-29)}} {2(1)}$
Which gives me 3.475 and -7.475. I'm not sure where I went wrong, any help would be greatly appreciated. Neither one of there are in my interval and my solution says it needs to be 4.48.

2. Apr 6, 2017

### MAGNIBORO

look the equation:
$$x = \frac {-2 \pm \sqrt{2^2 -(4)(1)(-29)}} {2(1)}$$
and solve it carefully

3. Apr 6, 2017

### Schaus

I swear I put that in my calculator a few times but this last time it worked... Thanks a lot anyways :)