Question

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?

Answer 1

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

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 \notin [1,4]`.

As for the derivative, using the ratio formula

`d/dx f(x)/g(x) = \frac{f'(x) g(x) - f(x) g'(x)}{g^2(x)}`

we have that `d/dx x/{x+2} = 2/{(x+2)^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 \in [1,4]` such that

`f'(c) = \frac{f(4)-f(1)}{4-1}`

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

`2/{(c+2)^2} = 1/3 (2/3-1/3)=1/9`

Isolating the terms involving `c`, we get

`(c+2)^2 = 18 \implies c+2=\pm\sqrt(18) \implies c=\pm\sqrt(18)-2`

Since `-sqrt(18)-2 = 6.24...`, we can't accept this solution, while `sqrt(18)-2 = 2.24....`, and it's thus ok.

As you can see in this link, the derivative evaluated in that point equals to `1/9`, just as we wanted.