(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then.
Differentiability Analysis of Function $f(x)$
The function is given by $f(x) = |\log_e x| - |x - 1|$ for $x \in (0, \infty)$. We analyze its differentiability, focusing on the point $x=1$ where the absolute value arguments become zero.
First, we express $f(x)$ piecewise:
- For $0 < x < 1$: $\log_e x < 0$ and $x - 1 < 0$. So, $f(x) = -(\log_e x) - (-(x - 1)) = x - \log_e x - 1$.
- For $x = 1$: $f(1) = |\log_e 1| - |1 - 1| = |0| - |0| = 0$.
- For $x > 1$: $\log_e x > 0$ and $x - 1 > 0$. So, $f(x) = (\log_e x) - (x - 1) = \log_e x - x + 1$.
Statement (I): $f$ is differentiable at all $x > 0$.
We check differentiability at $x=1$ by comparing the left-hand derivative (LHD) and the right-hand derivative (RHD).
LHD at $x=1$:
Using the derivative for $0 < x < 1$, $f'(x) = 1 - \frac{1}{x}$.
LHD = $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} \left(1 - \frac{1}{x}\right) = 1 - 1 = 0$.
RHD at $x=1$:
Using the derivative for $x > 1$, $f'(x) = \frac{1}{x} - 1$.
RHD = $\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} \left(\frac{1}{x} - 1\right) = 1 - 1 = 0$.
Since LHD = RHD = 0, the function $f$ is differentiable at $x=1$. For $x \neq 1$, the function is composed of standard functions whose absolute values are differentiable away from zero. Therefore, $f$ is differentiable for all $x > 0$.
Conclusion for Statement (I): Statement (I) is TRUE.
Monotonicity Analysis in $(0, 1)$
Statement (II): $f$ is increasing in $(0, 1)$.
For $0 < x < 1$, $f(x) = x - \log_e x - 1$.
The derivative is $f'(x) = 1 - \frac{1}{x}$.
In the interval $(0, 1)$, we have $0 < x < 1$, which means $\frac{1}{x} > 1$.
Therefore, $f'(x) = 1 - \frac{1}{x} < 0$ for $x \in (0, 1)$.
A negative derivative indicates the function is decreasing.
Conclusion for Statement (II): Statement (II) is FALSE.
Monotonicity Analysis in $(1, \infty)$
Statement (III): $f$ is decreasing in $(1, \infty)$.
For $x > 1$, $f(x) = \log_e x - x + 1$.
The derivative is $f'(x) = \frac{1}{x} - 1$.
In the interval $(1, \infty)$, we have $x > 1$, which means $0 < \frac{1}{x} < 1$.
Therefore, $f'(x) = \frac{1}{x} - 1 < 0$ for $x \in (1, \infty)$.
A negative derivative indicates the function is decreasing.
Conclusion for Statement (III): Statement (III) is TRUE.
Overall Conclusion
Summarizing the findings:
- Statement (I) is TRUE.
- Statement (II) is FALSE.
- Statement (III) is TRUE.
Thus, only statements (I) and (III) are TRUE.
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