Consider the following three statements for t | vedclass

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(D) The function is $f(x) = |\log_{e} x| - |x - 1|$.For $x \in (0, 1)$, $\log_{e} x < 0$ and $x - 1 < 0$, so $f(x) = -\log_{e} x - (-(x - 1)) = -\log_

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Only $(I)$ and $(III)$ are $TRUE$.

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(D) The function is $f(x) = |\log_{e} x| - |x - 1|$.For $x \in (0, 1)$, $\log_{e} x Then $f'(x) = -\frac{1}{x} + 1 = \frac{x - 1}{x}$. Since $x \in (0, 1)$, $f'(x) For $x \in (1, \infty)$, $\log_{e} x > 0$ and $x - 1 > 0$, so $f(x) = \log_{e} x - (x - 1) = \log_{e} x - x + 1$.Then $f'(x) = \frac{1}{x} - 1 = \frac{1 - x}{x}$. Since $x > 1$, $f'(x) At $x = 1$, $f(x) = |\log_{e} x| - |x - 1|$. The left-hand derivative at $x = 1$ is $\lim_{h 	o 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^-} \frac{-\log_{e}(1+h) + h}{h} = -1 + 1 = 0$. The right-hand derivative at $x = 1$ is $\lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^+} \frac{\log_{e}(1+h) - h}{h} = 1 - 1 = 0$. Since $f'(1) = 0$, the function is differentiable at $x = 1$. However, at $x = 1$, the absolute value functions $|\log_{e} x|$ and $|x - 1|$ are differentiable. Checking $x=1$ specifically, $f$ is differentiable. But for $f$ to be differentiable at all $x > 0$, we check $x=1$. Since $f'(1)=0$, it is differentiable. However, the statement $(I)$ is $TRUE$. Thus, only $(I)$ and $(III)$ are $TRUE$.

Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_{e} x| - |x - 1|$:
$(I)$ $f$ is differentiable at all $x > 0$.
$(II)$ $f$ is increasing in $(0, 1)$.
$(III)$ $f$ is decreasing in $(1, \infty)$.
Then:

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Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_{e} x| - |x - 1|$:$(I)$ $f$ is differentiable at all $x > 0$.$(II)$ $f$ is increasing in $(0, 1)$.$(III)$ $f$ is decreasing in $(1, \infty)$.Then: