Properties and Extremum Analysis of Logarithmic Radical Function

To determine which statement is true for the function $f: (0, \infty) \rightarrow (-\infty, \infty)$ defined by $f(x) = \sqrt{x} \log_e(x) - x + 1,$ we analyze its first and second derivatives.

Step 1: Derivative Analysis

First, we find the first derivative $f'(x)$ for $x \in (0, \infty)$: $f'(x) = \frac{d}{dx} \left( x^{1/2} \log_e(x) - x + 1 \right)$ $f'(x) = \frac{1}{2\sqrt{x}} \log_e(x) + \sqrt{x} \cdot \frac{1}{x} - 1$ $f'(x) = \frac{\log_e(x) + 2}{2\sqrt{x}} - 1$

Next, we calculate the second derivative $f''(x)$: $f''(x) = \frac{d}{dx} \left( \frac{1}{2} x^{-1/2} \log_e(x) + x^{-1/2} - 1 \right)$ $f''(x) = \frac{1}{2} \left( -\frac{1}{2} x^{-3/2} \log_e(x) + x^{-1/2} \cdot \frac{1}{x} \right) - \frac{1}{2} x^{-3/2}$ $f''(x) = -\frac{\log_e(x)}{4 x^{3/2}} + \frac{1}{2 x^{3/2}} - \frac{1}{2 x^{3/2}}$ $f''(x) = -\frac{\log_e(x)}{4 x^{3/2}}$

Step 2: Evaluating Monotonicity of $f'(x)$

Let us check the sign of $f''(x)$ in different intervals:

1. For $x \in (0, 1)$, we have $\log_e(x) < 0$, which implies: $f''(x) = -\frac{\log_e(x)}{4 x^{3/2}} > 0$ Since $f''(x) > 0$ for $x \in (0, 1)$, the derivative $f'(x)$ is strictly increasing on $(0, 1)$. Thus, option (A) is false.

2. For $x \in (1, \infty)$, we have $\log_e(x) > 0$, which implies: $f''(x) = -\frac{\log_e(x)}{4 x^{3/2}} < 0$ Thus, $f'(x)$ is strictly decreasing on $(1, \infty)$.

Step 3: Determining Local Extrema of $f(x)$

Since $f'(x)$ increases on $(0, 1]$ and decreases on $[1, \infty)$, the derivative $f'(x)$ achieves its maximum value on $(0, \infty)$ at $x = 1$.

Evaluating $f'(1)$: $f'(1) = \frac{\log_e(1) + 2}{2\sqrt{1}} - 1 = \frac{0 + 2}{2} - 1 = 0$

Since $x = 1$ is the absolute maximum point of $f'(x)$ and $f'(1) = 0$, it follows that: $f'(x) < 0 \quad \text{for all } x \in (0, 1) \cup (1, \infty)$

Since $f'(x) \le 0$ for all $x \in (0, \infty)$ and $f'(x) = 0$ only at the isolated point $x = 1$, $f'(x)$ does not change sign across $x = 1$. Consequently, the function $f(x)$ is strictly decreasing on the entire domain $(0, \infty)$.

Therefore, the function $f$ has neither a point of local maximum nor a point of local minimum in the interval $(0, \infty)$.

Conclusion

The correct option is D.