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.