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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) = 2x – x2 on [0, 1] |
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Answer» Given as f(x) = 2x – x2 on [0, 1] 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 [0, 1] and differentiable in (0, 1). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied. So, there exist a point c ∈ (0, 1) such that: f'(c) = (f(1) - f(0))/(1 - 0) ⇒ f’(c) = f(1) – f(0) f (x) = 2x – x2 Differentiate with respect to x: f’(x) = 2 – 2x For the f’(c), put the value of x = c in f’(x): f’(c)= 2 – 2c For the f(1), put the value of x = 1 in f(x): f (1)= 2(1) – (1)2 = 2 – 1 = 1 For the f(0), put the value of x = 0 in f(x): f (0) = 2(0) – (0)2 = 0 – 0 = 0 f’(c) = f(1) – f(0) ⇒ 2 – 2c = 1 – 0 ⇒ – 2c = 1 – 2 ⇒ – 2c = – 1 c = -1/-2 = (1/2) ∈ (0,1) Thus, lagrange's mean value theorem is verified. |
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