Question

verify rolles theorem and determine c in f(x)=(x+2)(x+1)(x-3) in [2,3]

Answer 1

f(x) = (x + 2)(x + 1)(x - 3), \(x \in [2, 3]\)

∵ f(x) is a polynomial of degree 3

∴ f(x) is continuous and differentiable in every domain.

But f(2) = 4 × 3 × -1 = -12

and f(3) = 5 × 4 × 0 = 0

∵ f(2) ≠ f(3)

∴ Rolle's theorem is not applicable in [2, 3].

There is a one mistake

Let x ∈ [-2, 3].

Then f(-2) = 0 × -1 × -4 = 0 and f(3) = 5 × 4 × 0 = 0

∴ f(-2) = f(3).

Hence, Rolle's theorem is applicable for internal [-2, 3].

Then there exist c ∈ (-2, 3) such that f'(c) = 0.

\(\Rightarrow\) 3c² - 7 = 0 

(∵ f(x) = (x + 2) (x + 1)(x - 3)

= (x² + 3x + 2) (x - 3)

= x³ - 7x - 6

∴ f'(x) = 3x² - 7)

\(\Rightarrow\) c² = \(\frac{7}{3}\)

\(\Rightarrow\) \(c = \pm \sqrt{\frac{7}{3}}\) ∈ [-2, 3].

Hence, Rolle's theorem is verified.