Verify Mean Value Theorem, if f (x) = x3 – 5x2 – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f′(c) = 0.
Given: f(x) = x3 – 5x2 – 3x in the interval [1, 3]
Mean Value Theorem states that for a function f : [a, b] → R, if
(a)f is continuous on [a, b]
(b)f is differentiable on (a, b)
Then there exists some c ∈ (a, b) such that 
As f(x) is a polynomial function,
(a) f(x) is continuous in [1, 3]
(b) f'(x) = 3x2 - 10x - 3
So, f(x) is differentiable in (1, 3).
∴ 
f(3) = 33 – 5(3)2 – 3(3) = 27 – 45 - 9 = -27
f(1) = 13 – 5(1)2 – 3(1)= 1 – 5 - 3 = -7
⇒ 
∴ There is a point c ∈ (1, 4) such that f'(c) = -10
⇒ f'(c) = -10
⇒ 3c2 – 10c - 3 = -10
⇒ 3c2 – 10c +7 =0
⇒ 3c2 - 3c – 7c + 7 = 0
⇒ 3c(c – 1) – 7(c – 1) = 0
⇒(c – 1)(3c – 7) = 0
⇒ c = 1, 7/3 where c = 7/3 ∈ (1, 3)
The Mean Value Theorem is verified for the given f(x) and c = 7/3 ∈ (1, 3) is the only point for which f'(c) = 0.