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Verify Lagrange’s mean value theorem for the functions on the indicated intervals. Find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem: f(x) = x(x – 1) on [1, 2] |
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Answer» Given as f(x) = x (x – 1) on [1, 2] = x2 – x The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. Therefore it is continuous in [1, 2] and differentiable in (1, 2). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied. So, there exist a point c ∈ (1, 2) such that: f'(c) = (f(2) - f(1))/(2 - 1) f'(c) = (f(2) - f(1))/1 f (x) = x2 – x Differentiate with respect to x f’(x) = 2x – 1 For the f’(c), put the value of x=c in f’(x): f’(c)= 2c – 1 For the f(2), put the value of x = 2 in f(x) f (2) = (2)2 – 2 = 4 – 2 = 2 For the f(1), put the value of x = 1 in f(x): f (1) = (1)2 – 1 = 1 – 1 = 0 ∴ f’(c) = f(2) – f(1) ⇒ 2c – 1 = 2 – 0 ⇒ 2c = 2 + 1 ⇒ 2c = 3 c = (3/2) ∈ (1,2) Thus, lagrange's mean value theorem is verified. |
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