Given the function #f(x)=x^(2/3)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [0,1] and find the c?

Question

Charles Anctil · Accepted Answer

#f# is the composition of functions that are continuus on #RR#, so it is continuous on #[0,1]#. #f'(x)# exists for all #x# other than #0#, so #f# is differentiable on #(0,1)# Therefore, #f# satisfies the hypotheses of the Mean Value theorem. Therefore there is a #c# in #(01,)# such that #f'(c) = (f(1)-f(0))/(1-0)# To actually find the #c#, solve the equation on the interval. You should get #c=8/27#

Ezra Adams · Answer

undefined To determine if $ f(x) = x^{2/3} $ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,1]$, we need to check if the function is continuous on the closed interval $[0,1]$ and differentiable on the open interval $(0,1)$.

$ f(x) = x^{2/3} $ is continuous for all real numbers. To check differentiability, we can compute its derivative.

$$ f'(x) = \frac{2}{3}x^{-1/3} $$

$ f'(x) $ exists for all $ x \neq 0 $, so it is differentiable on the interval $(0,1)$.

Therefore, since $ f(x) = x^{2/3} $ is continuous on $[0,1]$ and differentiable on $(0,1)$, it satisfies the hypotheses of the Mean Value Theorem.

To find the $ c $ value, we use the Mean Value Theorem formula:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

where $ a = 0 $, $ b = 1 $.

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

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

Now, we need to solve for $ c $ in $ f'(c) = 1 $.

$$ \frac{2}{3}c^{-1/3} = 1 $$

$$ c^{-1/3} = \frac{3}{2} $$

$$ c = \left(\frac{2}{3}\right)^{-3} = \frac{8}{27} $$

So, $ c = \frac{8}{27} $.