Correct answer is B.
It shows that f(x) is continuous on 1, 3 and derivable on 1, 3.
So, both the conditions of Lagrange’s Theorem are satisfied.
Consequently, there exists c Є 1, 3 such that
x = \(\pm \sqrt{3}\) Hence, c = \(\sqrt{3}\) Є (1, 3) such that f'(c) = \(\frac{f(3) - f(1)}{3 - 1}.\)