How do you verify that the hypothesis of the Mean-Value Theorem are satisfied on the interval [2,5], and find all values of c in the given interval that satisfy the conclusion of the theorem for f(x) = 1 / (x-1)f(x)=1x1?

1 Answer
May 20, 2015

The function f(x)=1/(x-1)f(x)=1x1 is certainly continuous and differentiable when x!=1x1, so it is continuous on the closed interval [2,5][2,5] and differentiable on the open interval (2,5)(2,5). The hypotheses of the Mean Value Theorem are therefore satisfied.

Hence, the conclusion is true: there is at least one number c\in (2,5)c(2,5) such that f'(c)=(f(5)-f(2))/(5-2)=(1/4-1)/3=(-3/4)/3=-1/4.

To find the value(s) of c where this is true, we find f'(c)=-(c-1)^(-2)=(-1)/((c-1)^2) and set this equal to -1/4 to get (c-1)^2=4 so that c-1=\pm 2 and c=1\pm 2. Of these, only c=1+2=3 is in the interval (2,5), so this is the one value of c guaranteed to exist by the Mean Value Theorem in this situation.