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Exercise 2.3 · Q61

Q.Verify Lagrange's mean value theorem for the function f(x)=x−1x−3f(x) = \dfrac{x-1}{x-3} on [4,5][4, 5].

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f(x)=x−1x−3f(x)=\dfrac{x-1}{x-3} is continuous and differentiable on [4,5][4,5] (the excluded point x=3x=3 lies outside this interval).

f(4)=31=3f(4)=\dfrac{3}{1}=3. f(5)=42=2f(5)=\dfrac{4}{2}=2. Chord slope =2−35−4=−1=\dfrac{2-3}{5-4}=-1.

By the quotient rule: f′(x)=(1)(x−3)−(x−1)(1)(x−3)2=−2(x−3)2f'(x)=\dfrac{(1)(x-3)-(x-1)(1)}{(x-3)^2}=\dfrac{-2}{(x-3)^2}. …

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