Consider the function $f: (0, \infty) \rightarrow (-\infty, \infty)$ given by

$f(x) = \sqrt{x}\, \log_e(x) - x + 1$.

Then which one of the following statements is TRUE?

Given $f(x) = \sqrt{x} \log_e(x) - x + 1$

Differentiating with respect to $x$:

$f'(x) = \dfrac{1}{2\sqrt{x}} \log_e(x) + \sqrt{x} \left(\dfrac{1}{x}\right) - 1$

$f'(x) = \dfrac{\log_e(x)}{2\sqrt{x}} + \dfrac{1}{\sqrt{x}} - 1$

Differentiating again with respect to $x$:

$f''(x) = \dfrac{d}{dx} \left( \dfrac{1}{2} x^{-1/2} \log_e(x) + x^{-1/2} - 1 \right)$

$f''(x) = \dfrac{1}{2} \left( -\dfrac{1}{2} x^{-3/2} \log_e(x) + x^{-1/2} \cdot \dfrac{1}{x} \right) - \dfrac{1}{2} x^{-3/2}$

$f''(x) = -\dfrac{\log_e(x)}{4x^{3/2}} + \dfrac{1}{2x^{3/2}} - \dfrac{1}{2x^{3/2}}$

$f''(x) = -\dfrac{\log_e(x)}{4x^{3/2}}$

For $x \in (0, 1)$, $\log_e(x) < 0$, which implies $f''(x) > 0$. Thus, $f'(x)$ is strictly increasing in $(0, 1)$.

For $x \in (1, \infty)$, $\log_e(x) > 0$, which implies $f''(x) < 0$. Thus, $f'(x)$ is strictly decreasing in $(1, \infty)$.

Therefore, $f'(x)$ attains its maximum value at $x = 1$.

Maximum value of $f'(x) = f'(1) = \dfrac{\log_e(1)}{2} + 1 - 1 = 0$.

Since the maximum value of $f'(x)$ is $0$, we have $f'(x) \le 0$ for all $x \in (0, \infty)$, with equality holding only at $x = 1$.

This means $f(x)$ is a strictly decreasing function on $(0, \infty)$.

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

Answer: The function $f$ has NEITHER a point of local maximum NOR a point of local minimum in the interval $(0, \infty)$