How do you verify that the hypotheses of the Mean-Value Theorem are satisfied on the interval [-3,0] and then find all values of c in this interval that satisfy the conclusion of the theorem for # f(x) = 1/(x-1)#?

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Answer 1 · 2015-09-11T19:44:01

See the explanation.

#f# is continuous except at #x=1# which is not in #[-3,0]#, so #f# is continuous on #[-3,0]#

#f'(x) = -1/(x-1)^2# exists for all #x# except #x=1# which is not in #(-3,0)#, so #f# is differentiable on #(-3,0)#

Thus, the hypotheses are satisfied.

To finish the question, set #f'(x) = (f(0)-f(-3))/(0-(-3))# and solve for solutions in #(-3,0)#.

#-1/(x-1)^2 = -1/4# implies #x = 3, -1#.
The first is not in the interval, so it is not the #c# referred to in the conclusion.

#c=-1#