Concept:If real valued function f (x)(i) is continuous in [a, b](ii) is differentiable on (a, b)(iii) f'(c) = \(\rm\frac{[f (b) - f(a)] }{(b - a)}\) Calculation:(i) is continuous in [a, b](ii) is differentiable on (a, b)f (x) = x (x - 4)f (a) = f (1) = 1 \(\times\) (1 - 4) = 1 \(\times\) (-3) = -3f (b) = f (4) = 4 \(\times\) (4 - 4) = 4 \(\times\) 0 = 0f'(c) = \(\rm\frac{[f (b) - f(a)] }{(b - a)}\) = \(\rm\frac{[0 -(-3)] }{(4 - 1)}\) = \(\rm\frac{3}{3}\) = 1As per the mean value theorem statement, there is a point c ? (1, 4) such that f'(c) = [f (b) “ f (a)]/ (b “ a), i.e. f'(c) = 1.2c - 4 = 12c = 1 + 4c = 5/2