Question

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)`?

Answer 1

The function `f(x)=1/(x-1)` is certainly continuous and differentiable when `x!=1`, so it is continuous on the closed interval `[2,5]` and differentiable on the open interval `(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)` 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.