Question

f(x) = x³ – 2x² – x + 3 in [0, 1]

Answer 1

We have, f(x) = x³ – 2x² – x + 3 in [0, 1]

Since, f(x) is a polynomial function it is continuous in [0, 1] and differentiable in (0, 1)

Thus, conditions of mean value theorem are satisfied.

Hence, there exists a real number c ∈ (0, 1) such that

f'(c) = `("f"(1) - "f"(0))/(1 - 0)`

⇒ 3c² – 4c – 1 = `([1 - 2 - 1 + 3] - [0 + 3])/(1 - 0)`

⇒ 3c² – 4c – 1 = –2

⇒ 3c² – 4c + 1 = 0

⇒ (3c – 1)(c – 1) = 0

⇒ c = `1/3 ∈ (0, 1)`

Hence, the mean value theorem has been verified.