Given the function #f(x)=sqrt(2-x)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-7,2] and find the c?

Question

Cameron Alexander · Accepted Answer

You determine whether it satisfies the hypotheses by determining whether #f(x) = sqrt(2-x)# is continuous on the interval #[-7,2]# and differentiable on the interval #(-7,2)#.  (Those are the hypotheses of the Mean Value Theorem) You find the #c# mentioned in the conclusion of the theorem by solving #f'(x) = (f(2)-f(-7))/(2-(-7))# on the interval #(-7,2)#. #f# is continuous on its domain, which includes #[-7,2]# #f'(x) = (-1)/(2sqrt(2-x))# which exists for all #x < 2# so it exists for all #x# in #(-7,2)# Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval. To find #c# solve the equation #f'(x) = (f(2)-f(-7))/(2-(-7))#.  Discard any solutions outside #(-7,2)#. You should get #c = -1/4#.

Emilia Aguilar · Answer

undefined To determine whether the function $ f(x) = \sqrt{2 - x} $ satisfies the hypotheses of the Mean Value Theorem on the interval $[-7, 2]$, we need to check if the following conditions are met:

1. $ f(x) $ is continuous on $[-7, 2]$.
2. $ f(x) $ is differentiable on $(-7, 2)$.

$ f(x) = \sqrt{2 - x} $ is continuous everywhere except at $ x = 2 $ because the square root function is continuous for all real numbers except when its argument is negative.

To check differentiability, we need to ensure that the derivative exists for all $ x $ in $(-7, 2)$. Taking the derivative of $ f(x) $:

$$ f'(x) = \frac{d}{dx} \sqrt{2 - x} = \frac{1}{2\sqrt{2 - x}} $$

$ f'(x) $ exists for all $ x $ in $(-7, 2)$ except at $ x = 2 $ where $ f(x) $ is not differentiable. However, this point is not in the interval $[-7, 2]$, so it does not affect the applicability of the Mean Value Theorem on this interval.

Since both conditions are satisfied, we can apply the Mean Value Theorem. According to the theorem, there exists a point $ c $ in $(-7, 2)$ such that:

$$ f'(c) = \frac{f(2) - f(-7)}{2 - (-7)} $$

$$ f'(c) = \frac{\sqrt{2 - 2} - \sqrt{2 - (-7)}}{2 - (-7)} $$

$$ f'(c) = \frac{\sqrt{0} - \sqrt{9}}{9} $$

$$ f'(c) = \frac{0 - 3}{9} $$

$$ f'(c) = -\frac{1}{3} $$

Thus, there exists $ c $ in $(-7, 2)$ such that $ f'(c) = -\frac{1}{3} $.