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Verify Rolle’s Theorem for the function: f(x) = (x – 2)4 (x – 3)3 in [2, 3] |
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Answer» We know that (i) f (x) = (x – 2)4 (x – 3)3 is a polynomial which is continuous for all x ϵ R Hence, f (x) = (x – 2)4 (x – 3)3 is continuous on [2, 3] (ii) f’(x) = 4 (x – 2)3 (x – 3)3 + 3 (x – 2)4 (x – 3)2 exist in [2, 3] Hence, f (x) = (x – 2)4 (x – 3)3 is differentiable on (2, 3) (iii) We know that f (2) = (2 – 2)4 (2 – 3)3 = 0 Similarly f (3) = (3 – 2)4 (3 – 3)3 = 0 Here f (2) = f (3) The conditions of Rolle’s Theorem are satisfied. There exist at least one c ϵ (2, 3) where f’(c) = 0 So we get 4 (c – 2)3 (c – 3)3 + 3 (c – 2)4 (c – 3)2 = 0 We know that (c – 2)3 (c – 3)2 (7c – 18) = 0 Which gives c = 2 or 3 or 18/7 We know that value of c = 18/7 ϵ (2, 3) Therefore, Rolle’s Theorem is satisfied. |
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