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