Given the function #f(x)=(x+1)/x#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1/2, 2] and find the c?

Question

Ezekiel Alexander · Accepted Answer

Please see below.

Hypothesis : #f# is continuous on #[1/2,2]# True. #f# is a rational function, so #f# is continuous on its domain. The domain is all reals except #0# and #0# is not in #[1/2,2]#. Hypothesis: #f# is differentiable on #(1/2,2)# True. Remember that "differentiable" means "the derivative exists". We have #f'(x)=-1/x^2#, so #f'(x)# exists for all reals except #0# and #0# is not in #(1/2,2)#. The conclusion says that there is a #c# in #(1/2,2)# with #f'(c) = (f(2)-f(1/2))/(2-(1/2))#. To find those #c#'s, solve the equation. Discard solutions outside of #(1/2,2)#. #-1/x^2 = -1# has solutions #x = +-1#. So #c = 1#

Christopher Agresta · Answer

undefined To determine if the function satisfies the hypotheses of the Mean Value Theorem on the interval [1/2, 2], we need to check two conditions:

1. The function must be continuous on the closed interval [1/2, 2].
2. The function must be differentiable on the open interval (1/2, 2).

Given the function $ f(x) = \frac{x+1}{x} $, it is continuous and differentiable for all $ x $ except $ x = 0 $ (where it's not defined). Thus, it satisfies the first condition.

To check the differentiability condition, we find the derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx}\left(\frac{x+1}{x}\right) = \frac{d}{dx}\left(1 + \frac{1}{x}\right) = -\frac{1}{x^2} $$

The function $ f'(x) = -\frac{1}{x^2} $ is defined and continuous for all $ x $ in the interval [1/2, 2], so it satisfies the second condition.

Since the function $ f(x) = \frac{x+1}{x} $ satisfies both conditions, the Mean Value Theorem applies.

Now, to find the value of $ c $, we use the Mean Value Theorem formula:
$$ f'(c) = \frac{f(2) - f(1/2)}{2 - 1/2} $$

First, find $ f(2) $ and $ f(1/2) $:
$$ f(2) = \frac{2 + 1}{2} = \frac{3}{2} $$
$$ f(1/2) = \frac{1/2 + 1}{1/2} = \frac{3}{2} $$

$$ f(2) - f(1/2) = \frac{3}{2} - \frac{3}{2} = 0 $$

$$ f'(c) = \frac{0}{2 - 1/2} = 0 $$

Now, solve for $ c $:
$$ -\frac{1}{c^2} = 0 $$
$$ c^2 = \infty $$
$$ c = \infty $$

So, there is no specific value of $ c $ in the interval (1/2, 2) where the derivative of the function is equal to the average rate of change of the function over the interval. Therefore, the conclusion drawn from the Mean Value Theorem is that there exists a point $ c $ such that $ f'(c) = 0 $, but $ c $ is not in the interval (1/2, 2).