# How do I find the numbers c that satisfy the Mean Value Theorem for f(x)=x/(x+2) on the interval [1,4] ?

Sep 12, 2014

The mean value theorem guarantees that there exists a number $c$ in $\left(1 , 4\right)$ such that
$f ' \left(c\right) = \frac{f \left(4\right) - f \left(1\right)}{4 - 1}$.
The actual value of $c$ is $- 2 + 3 \sqrt{2}$.

Let us find the left-hand side of the above equation,
By Quotient Rule,
$f ' \left(x\right) = \frac{1 \cdot \left(x + 2\right) - x \cdot 1}{{\left(x + 2\right)}^{2}} = \frac{2}{x + 2} ^ 2$
$R i g h t a r r o w f ' \left(c\right) = \frac{2}{c + 2} ^ 2$

Let us find the right-hand side,
$\frac{f \left(4\right) - f \left(1\right)}{4 - 1} = \frac{\frac{4}{6} - \frac{1}{3}}{3} = \frac{1}{9}$

By setting the left-hand side and the right-hand side equal to each other,
$\frac{2}{c + 2} ^ 2 = \frac{1}{9}$

by taking the reciprocal,
${\left(c + 2\right)}^{2} / 2 = 9$

by multiplying by 2,
${\left(c + 2\right)}^{2} = 18$

by taking the square-root,
$c + 2 = \pm \sqrt{18} = \pm 3 \sqrt{2}$

by subtracting 2,
$c = - 2 \pm 3 \sqrt{2}$

Since $1 < c < 4$,
$c = - 2 + 3 \sqrt{2}$