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Exercise 2.3 · Q49

Q.Check the validity of the Rolle's theorem for the function f(x)=x2/3, x∈[−1,1]f(x) = x^{2/3},\ x \in [-1, 1].

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f(−1)=(−1)2/3=1f(-1)=(-1)^{2/3}=1. f(1)=12/3=1f(1)=1^{2/3}=1. So f(−1)=f(1)=1f(-1)=f(1)=1 — the endpoint condition holds.

f′(x)=23x−1/3=23x3f'(x)=\dfrac23x^{-1/3}=\dfrac{2}{3\sqrt[3]{x}}. At x=0x=0, this is undefined (division by zero), so ff fails to be differentiable at the interior point x=0∈(−1,1)x=0\in(-1,1). …

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