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?

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?

Christopher Agresta · Accepted Answer

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 
otin [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.

Christopher Agresta · Answer

undefined To verify that the function $ f(x) = \frac{x}{x+2} $ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,4]$, we first check if the function is continuous on the closed interval $[1,4]$ and differentiable on the open interval $(1,4)$. The function is continuous and differentiable on the given interval.

Next, we find the derivative of $ f(x) $ to apply the Mean Value Theorem. The derivative is given by:

$$ f'(x) = \frac{d}{dx}\left(\frac{x}{x+2}\right) = \frac{2}{(x+2)^2} $$

Now, we check if $ f'(x) $ is continuous on $[1,4]$, which it is. Therefore, all conditions of the Mean Value Theorem are satisfied.

To find all numbers $ c $ that satisfy the conclusion of the Mean Value Theorem, we apply the theorem:

$$ f'(c) = \frac{f(4) - f(1)}{4 - 1} $$
$$ f'(c) = \frac{\frac{4}{6} - \frac{1}{3}}{3} $$
$$ f'(c) = \frac{\frac{2}{3} - \frac{1}{3}}{3} $$
$$ f'(c) = \frac{1}{9} $$

Now, solve for $ c $:

$$ \frac{2}{(c+2)^2} = \frac{1}{9} $$
$$ 18 = (c+2)^2 $$
$$ c+2 = \pm \sqrt{18} $$
$$ c+2 = \pm 3\sqrt{2} $$
$$ c = -2 \pm 3\sqrt{2} $$

So, the numbers $ c $ that satisfy the conclusion of the Mean Value Theorem are $ c = -2 + 3\sqrt{2} $ and $ c = -2 - 3\sqrt{2} $.