Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem #f(x) = 1/(x-1)#; [-3, 0]?

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Answer 1 · 2016-12-14T07:28:31

There is only one number c #=-1#

The domain of #f(x)# is #D_f(x)=RR-{1}#

#f(x)# is continuos and defined on #[-3,0] # and differentiable on #] -3,0 [#, then there is #c# such that

#f'(c)=(f(0)-f(-3))/(0--3)=(f(0)-f(3))/3#

#f(0)=1/(0-1)=-1#

#f(-3)=1/(-3-1)=-1/4#

#f'(x)=-1/(x-1)^2#

#f'(c)=-1/(c-1)^2#

Therefore,

#-1/(c-1)^2=(-1+1/4)/(3)=-1/4#

So,

#(c-1)^2=4#

#c-1=+-2#

#c=3# or #c=-1#

We must have, #c in [-3,0] #

So, #c=-1#