# How do you verify that the function f(x) = (x)/(x+2) satisfies the hypotheses of the Mean Value Theorem on the given interval [1,4], then find all numbers c that satisfy the conclusion of the Mean Value Theorem?

Apr 6, 2015

The mean value theorem requires a function to be continuous in a closed interval $\left[a , b\right]$, and differentiable in the open interval $\left(a , b\right)$.

These conditions are easily checked, since the only point in which the function is not defined is $x = - 2$ (since in that point the denominator equals zero), and of course $- 2 \setminus \notin \left[1 , 4\right]$.

As for the derivative, using the ratio formula

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \setminus \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{g}^{2} \left(x\right)}$

we have that $\frac{d}{\mathrm{dx}} \frac{x}{x + 2} = \frac{2}{{\left(x + 2\right)}^{2}}$, and again the only point in which this function has no sense is $x = - 2$.

Now that we ensured ouselves to be in the right hypothesis, we must find a point $c \setminus \in \left[1 , 4\right]$ such that

$f ' \left(c\right) = \setminus \frac{f \left(4\right) - f \left(1\right)}{4 - 1}$

We know that f'(c) = 2/{(c+2)^2, and we can easily compute that $f \left(1\right) = \frac{1}{3}$ and $f \left(4\right) = \frac{2}{3}$. We can thus translate the previous equation into

$\frac{2}{{\left(c + 2\right)}^{2}} = \frac{1}{3} \left(\frac{2}{3} - \frac{1}{3}\right) = \frac{1}{9}$

Isolating the terms involving $c$, we get

${\left(c + 2\right)}^{2} = 18 \setminus \implies c + 2 = \setminus \pm \setminus \sqrt{18} \setminus \implies c = \setminus \pm \setminus \sqrt{18} - 2$

Since $- \sqrt{18} - 2 = 6.24 \ldots$, we can't accept this solution, while $\sqrt{18} - 2 = 2.24 \ldots .$, and it's thus ok.

As you can see in this link, the derivative evaluated in that point equals to $\frac{1}{9}$, just as we wanted.