How do you give a value of c that satisfies the conclusion of the Mean Value Theorem for Derivatives for the function #f(x)=-2x^2-x+2# on the interval [1,3]?

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Jim H Share
Mar 23, 2015

The conclusion of the Mean Value Theorem says that there is a number #c# in the interval #(1, 3)# such that:
#f'(c)=(f(3)-f(1))/(3-1)#

To find (or try to find) #c#, set up this equation and solve for #c#.
If there's more than one #c# make sure you get the one (or more) in the interval #(1, 3)#.

For #f(x)--2x^2-x+2#, we have
#f(1)=-1#, and #f(3)=-18-3+2=-19#
Also,
#f'(x)=-4x-1#.

So the #c# we're looking for satisfies:

#f'(c)=-4c-1=(f(3)-f(1))/(3-1)=(-19--1)/(3-1)=(-18)/2=-9#

So we need

#-4c-1=-9#. And #c=2#.

Note:
I hope you've been told that actually finding the value of #c# is not a part of the Mean Value Theorem.
The additional question"find the value of #c#" is intended as a review of your ability to solve equations. For most functions, you will not be able to find the #c# that the MVT guarantees us is there..

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