Given f(x) = 4 - (frac{1}{2} - x)^{2/3},g(x) | vedclass

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(A) Lagrange's Mean Value Theorem $(LMVT)$ requires a function to be continuous on $[a, b]$ and differentiable on $(a, b)$.$1$. For $f(x) = 4 - (\frac

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$f, g, h$

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(A) Lagrange's Mean Value Theorem $(LMVT)$ requires a function to be continuous on $[a, b]$ and differentiable on $(a, b)$.$1$. For $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$: The derivative $f'(x) = -\frac{2}{3}(\frac{1}{2} - x)^{-1/3}(-1) = \frac{2}{3(\frac{1}{2} - x)^{1/3}}$. Since $x = \frac{1}{2}$ lies in $(0, 1)$,$f'(x)$ is undefined at $x = \frac{1}{2}$. Thus,$f$ is not differentiable on $(0, 1)$.$2$. For $g(x)$: At $x = 0$,$\lim_{x 	o 0^+} g(x) = \lim_{x 	o 0^+} \frac{	an([x])}{x} = \frac{	an(0)}{0} = 0$,but $g(0) = 1$. Since $\lim_{x 	o 0^+} g(x) 
eq g(0)$,$g$ is not continuous at $x = 0$. Thus,$g$ is not continuous on $[0, 1]$.$3$. For $h(x) = \{x\}$: The fractional part function is discontinuous at every integer. In $[0, 1]$,it is discontinuous at $x = 0$ and $x = 1$. Thus,$h$ is not continuous on $[0, 1]$.$4$. For $k(x) = 5^{\log_2(x + 3)} = (x + 3)^{\log_2 5}$: This is a power function with a positive base for $x \in [0, 1]$. It is continuous and differentiable on $[0, 1]$.Therefore,$LMVT$ is not applicable to $f, g,$ and $h$.

List-$I$	List-$II$
$A. \|x-1\|$	$I. 2 \log (e^3+e^2)$
$B. \log x$	$II. 2$
$C. x^2+x+1$	$III. \log_3 e^2$
$D. e^x$	$IV. \sqrt{2}$
	$V. \log \left(\frac{e^3-e}{2}\right)$

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

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