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Show that the Lagrange's mean value theorem is not applicable to the function f(x) = 1/x on [–1, 1]. |
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Answer» Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that f(b) − f(a) = f′(c)(b − a) \(\Rightarrow f'(c) = \frac{f(b) - f(a)}{b - a}\) This theorem is also known as First Mean Value Theorem. f(x) = \(\frac{1}{x}\) on [-1, 1] Here, x ≠ 0 ⇒ f(x) exists for all values of x except 0 ⇒ f(x) is discontinuous at x = 0 ∴ f(x) is not continuous in [– 1, 1] Hence the lagrange’s mean value theorem is not applicable to the function f(x) = \(\frac{1}{x}\) on [-1, 1] |
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