Consider the function f:(0,2) rightarrow R de | vedclass

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(A) Given $f(x) = \frac{x}{2} + \frac{2}{x}$ for $x \in (0, 2)$.$f'(x) = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2}$.Since $x \in (0, 2)$,$x^2

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$g$ is continuous but not differentiable at $x=1$

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(A) Given $f(x) = \frac{x}{2} + \frac{2}{x}$ for $x \in (0, 2)$.$f'(x) = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2}$.Since $x \in (0, 2)$,$x^2 For $0 At $x=1$,$g(1) = \frac{1}{2} + \frac{2}{1} = \frac{5}{2}$.For $1 Since $\lim_{x ightarrow 1^-} g(x) = \lim_{x ightarrow 1^+} g(x) = g(1) = \frac{5}{2}$,$g(x)$ is continuous at $x=1$.Now check differentiability at $x=1$:Left-hand derivative: $g'(1^-) = f'(1) = \frac{1}{2} - \frac{2}{1^2} = \frac{1}{2} - 2 = -\frac{3}{2}$.Right-hand derivative: $g'(1^+) = \frac{d}{dx}(\frac{3}{2} + x) = 1$.Since $g'(1^-) 
eq g'(1^+)$,$g(x)$ is not differentiable at $x=1$.

Consider the function $f:(0,2) \rightarrow R$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $g(x)=\begin{cases} \min \{f(t) : 0 < t \leq x\}, & 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{cases}$. Then,

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