Question

(D) The function \(f\) has NEITHER a point of local maximum nor a point of local minimum.

Step 1: Find the first derivative

To determine local extrema, first find the critical points of \(f(x)\) by computing its first derivative, \(f'(x)\).

\[ f(x)=\sqrt{x}\,\log_e(x) \]

Applying the product rule,

\[ f'(x) =\frac{1}{2\sqrt{x}}\log_e(x)+\sqrt{x}\cdot\frac{1}{x} =\frac{\log_e(x)+2}{2\sqrt{x}} \]

Step 2: Find the second derivative

To study the monotonicity of \(f'(x)\) and test statement (A), compute the second derivative.

\[ f''(x) = -\frac{\log_e(x)}{4x^{3/2}} \]

Step 3: Analyze intervals for the derivatives

Examine the sign of \(f''(x)\) to understand the behavior of \(f'(x)\).

For the interval \((0,1)\):

Since \[ \log_e(x)<0, \] we have \[ -\log_e(x)>0. \] Therefore, \[ f''(x)>0. \] Hence \(f'(x)\) is strictly increasing on \((0,1)\). This shows statement (A) is false.

At \(x=1\):

\[ f''(1)=0. \]

For the interval \((1,\infty)\):

Since \[ \log_e(x)>0, \] we have \[ -\log_e(x)<0. \] Therefore, \[ f''(x)<0. \] Hence \(f'(x)\) is strictly decreasing on \((1,\infty)\).

Step 4: Determine local extrema

Since \(f'(x)\) increases on \((0,1)\) and decreases on \((1,\infty)\), its maximum value occurs at \(x=1\).

\[ f'(1) = \frac{\log_e(1)+2}{2} = 1 \]

Now solve

\[ f'(x)=0 \]

\[ \frac{\log_e(x)+2}{2\sqrt{x}}=0 \]

\[ \log_e(x)=-2 \]

\[ x=e^{-2}. \]

Observe the sign change:

\[ f'(x)<0 \quad\text{for}\quad 0<x<e^{-2} \]

\[ f'(x)>0 \quad\text{for}\quad x>e^{-2} \]

Thus \(f'(x)\) changes from negative to positive at \(x=e^{-2}\), so \(f\) has a local minimum at

\[ x=e^{-2}. \]

Therefore the statement

(D) The function has neither a local maximum nor a local minimum

is false.