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Miscellaneous Exercise 2(II) · Q115

Q.Verify Rolle's theorem for the function f(x)=2ex+e−xf(x) = \dfrac{2}{e^x + e^{-x}} on [−1,1][-1, 1].

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Since ex+e−xe^x+e^{-x} is an even function of xx (unchanged when x→−xx\to-x), f(x)=2ex+e−xf(x)=\dfrac{2}{e^x+e^{-x}} is also even, so f(−1)=f(1)f(-1)=f(1) automatically. Both ff and its denominator (never zero, since ex+e−x>0e^x+e^{-x}>0 always) are continuous and differentiable everywhere, so Rolle's theorem applies.

f(x)=2(ex+e−x)−1f(x)=2(e^x+e^{-x})^{-1}. By the chain rule: f′(x)=−2(ex+e−x)−2(ex−e−x)=−2(ex−e−x)(ex+e−x)2f'(x)=-2(e^x+e^{-x})^{-2}(e^x-e^{-x})=\dfrac{-2(e^x-e^{-x})}{(e^x+e^{-x})^2}. …

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