A function f is defined by f(x)=2+(x-1)^{2/3} | vedclass

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(D) Given the function $f(x) = 2 + (x - 1)^{2/3}$ on the interval $[0, 2]$.First, we check the continuity: The function $(x - 1)^{2/3}$ is continuous

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Rolle's theorem is applicable on $[0,2]$

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(D) Given the function $f(x) = 2 + (x - 1)^{2/3}$ on the interval $[0, 2]$.First, we check the continuity: The function $(x - 1)^{2/3}$ is continuous for all $x \in \mathbb{R}$, so $f(x)$ is continuous on $[0, 2]$.Next, we check the differentiability: $f'(x) = \frac{2}{3}(x - 1)^{-1/3} = \frac{2}{3(x - 1)^{1/3}}$.At $x = 1$, $f'(x)$ is undefined because the denominator becomes zero. Thus, $f$ is not differentiable at $x = 1$, which lies in the interval $(0, 2)$.Since $f$ is not differentiable on $(0, 2)$, Rolle's theorem is not applicable on $[0, 2]$.Checking the values at endpoints: $f(0) = 2 + (0 - 1)^{2/3} = 2 + 1 = 3$ and $f(2) = 2 + (2 - 1)^{2/3} = 2 + 1 = 3$. Thus, $f(0) = f(2)$.Therefore, the statement '$f$ is not derivable in $(0, 2)$' is true,'$f$ is continuous in $[0, 2]$' is true,'$f(0) = f(2)$' is true, and 'Rolle's theorem is applicable on $[0, 2]$' is incorrect.

$A$ function $f$ is defined by $f(x)=2+(x-1)^{2/3}$ on $[0,2]$. Which of the following statements is incorrect?

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