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

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Answer 1 · 2016-10-01T02:34:01

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# , #a=-1# and #b=1#.

This function is a polynomial, so it is continuous on its domain, #(-oo, oo)#. In particular, #f# is continuous on #[-1, 1]#

#f'(x)= 2x# which exists for all real #x#, it exists for all #x# in #(-1,1)#,
So #f# is differentiable on #(-1, 1)#

So, both hypotheses are satisfied.

Because the hypotheses are satisfied, we can be sure that there is a #c# (at least one #c#) in #(a,b)# with #f'(c) = (f(b)-f(a))/(b-a)#.

To find the #c# (or #c#'s) mentioned in the conclusion, solve

#f'(x) = (f(1)-f(-1))/(1-(-1))#.

If there are solutions outside the interval #(-1,1)#, they are not values of #c# mentioned in the conclusion of MVT.

In this case the only solution is #0#, so #c=0#.