Discussion
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 ?
Answer: D
The true statement is that \(f\) has neither a local maximum nor a local minimum in \((0, \infty)\).
Differentiate and put everything over a common denominator:
\[ \begin{aligned} f'(x) &= \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}} - 1 \\ &= \frac{\ln x + 2 - 2\sqrt{x}}{2\sqrt{x}} \end{aligned} \]
Call the numerator \(g(x) = \ln x + 2 - 2\sqrt{x}\). Then \(g'(x) = \frac{1}{x} - \frac{1}{\sqrt{x}} = \frac{1 - \sqrt{x}}{x}\), so \(g\) rises on \((0, 1)\), falls after 1, and its largest value is \(g(1) = 0\). Hence \(f'(x) \le 0\) everywhere, with equality only at \(x = 1\): \(f\) is strictly decreasing and has no local extremum, which rules out the maximum and minimum options.
The first option fails too. Differentiating again gives \(f''(x) = -\frac{\ln x}{4x^{3/2}}\), which is positive on \((0, 1)\), so \(f'\) is increasing there, climbing from \(-\infty\) up to 0 at \(x = 1\).
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