f(x) = 1/x and its derivative f '(x) = -1/x² are discontinuous only at x = 0; everywhere else it is defined and behaves nicely in the sense that f is
• continuous on the closed interval [1, 3], and
• differentiable on the open interval (1, 3)
and the MVT holds.
The theorem says there is some real number c between 1 and 3 such that
[tex]f'(c) = \dfrac{f(3) - f(1)}{3 - 1}[/tex]
Solve for this c :
[tex]-\dfrac1{c^2} = \dfrac{\frac13-\frac11}{3 - 1} \implies c^2 = 3 \implies \boxed{c = \sqrt3}[/tex]
