`f(x) = 1/x, [1,3]` Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers `c` that satisfy the conclusion of the Mean Value Theorem.
It is sufficient for f to be continuous on [1, 3] and differentiable on (1, 3). This is obviously true (because 0, the only gap of f, isn't in this interval).
Then by the Mean Value theorem there is c on (1, 3) such that f'(c)=[f(3)-f(1)]/[3-1] = (1/3 - 1)/2 = -1/3. Let's find c.
`f'(x)=-1/(x^2)` and must be = -1/3. Then `x^2 = 3,` `x=+-sqrt(3).` `-sqrt(3)lt0` and is not in (1, 3) while c= `sqrt(3)` is.