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 [-1,8] and find the c?

1 Answer
Oct 12, 2016

The Mean Value Theorem has two hypotheses:

H1 : #f# is continuous on the closed interval #[a,b]#

H2 : #f# is differentiable on the open interval #(a,b)#.

In this question, #f(x) = x^(2/3)# , #a=-1# and #b=8#.

This function is continuous on its domain, #(-oo, oo)#, so it is continuous on #[-1, 8]#

#f'(x)=2/(3root(3)x)# which exists for all #x != 0#. But #0# is in #(-1,8)#

So #f# is not differentiable on #(-1, 8)#

To determine whether there is a #c# that satisfies the conclusion, we need to determine whether there is a #c# in (-1,8)# for which

#f'(c) = (f(8)-f(-1))/(8-(-1)) = (4-1)/9 = 1/3#

We can actually solve #2/(3root(3)x) = 1/3#.

We get #x=8# which is not in the open interval #(-1,8)#. So there is no #c# that satisfies the conclusion of MVT.

For an example of a function on an interval where it does not satisfy the hypotheses, but does satisfy the conclusion, please see:
https://socratic.org/questions/given-the-function-f-x-x-1-3-how-do-you-determine-whether-f-satisfies-the-hypoth