Monotonicity & Extrema
DAILY_CHALLENGE
Grade None
Question:
Consider the function $f:(0,\infty) \to (-\infty,\infty)$ given by
$$f(x) = \sqrt{x}\,\log_e(x) - x + 1.$$
Then which one of the following statements is TRUE?
The derivative of the function $f$ is decreasing in the interval $(0,1)$
The function $f$ has a local maximum at some point $a \in (0,\infty)$
The function $f$ has a local minimum at some point $b \in (0,\infty)$
The function $f$ has NEITHER a point of local maximum NOR a point of local minimum in the interval $(0,\infty)$
Step-by-Step Solution
Key Concept: A function with $f'(x) \le 0$ everywhere (with equality at only isolated points) is strictly monotone decreasing and has no local maxima or minima.
**Step 1: Compute the first derivative**
$f'(x) = \dfrac{\ln x}{2\sqrt{x}} + \dfrac{1}{\sqrt{x}} - 1 = \dfrac{\ln x + 2 - 2\sqrt{x}}{2\sqrt{x}}.$
**Step 2: Analyze the numerator $g(x) = \ln x + 2 - 2\sqrt{x}$**
$g'(x) = \dfrac{1}{x} - \dfrac{1}{\sqrt{x}} = \dfrac{1-\sqrt{x}}{x}$. So $g$ increases on $(0,1)$ and decreases on $(1,\infty)$, with maximum $g(1) = 0+2-2 = 0$. Hence $g(x) \le 0$ for all $x > 0$, with equality only at $x = 1$.
**Step 3: Conclude monotonicity of $f$**
Since $g(x) \le 0$ and $2\sqrt{x} > 0$, we get $f'(x) \le 0$ everywhere with equality only at $x=1$. Thus $f$ is strictly decreasing and has no local extrema.
Correct Answer: D