Question

Let f(x) = x² logx, x > 0. Then the minimum value of f is

• A. \(\frac{1}{\sqrt{e}}\)
• B. 2e
• C. -2e
• D. √e
• E. \(\frac{-1}{2e}\)

Answer 1

Answer: (E)

We are given the function \( f(x) = x^2 \log x \), where \( x > 0 \), and we are asked to find the minimum value of \( f(x) \).

To find the minimum value, we first compute the derivative of \( f(x) \) using the product rule. The function is a product of \( x^2 \) and \( \log x \), so we differentiate each part: 

Let \( u = x^2 \) and \( v = \log x \). Then, using the product rule:

\( f'(x) = u'v + uv' \),

where \( u' = 2x \) and \( v' = \frac{1}{x} \).

Thus, the derivative is:

\( f'(x) = 2x \log x + x^2 \cdot \frac{1}{x} = 2x \log x + x \).

Now, set \( f'(x) = 0 \) to find the critical points:

\( 2x \log x + x = 0 \),

\( x(2 \log x + 1) = 0 \).

Since \( x > 0 \), we must have \( 2 \log x + 1 = 0 \),

which simplifies to:

\( \log x = -\frac{1}{2} \),

so:

\( x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \).

This is the critical point. To determine if this is a minimum, we compute the second derivative of \( f(x) \):

\( f''(x) = \frac{d}{dx}\left(2x \log x + x\right) = 2 \log x + 2 + 1 = 2 \log x + 3 \).

At \( x = \frac{1}{\sqrt{e}} \), we have:

\( \log\left(\frac{1}{\sqrt{e}}\right) = -\frac{1}{2} \),

so:

\( f''\left(\frac{1}{\sqrt{e}}\right) = 2 \left(-\frac{1}{2}\right) + 3 = 2 \),

which is positive, indicating that the critical point is a minimum.

Thus, the minimum value of \( f(x) \) occurs at \( x = \frac{1}{\sqrt{e}} \). To find the minimum value of \( f(x) \), substitute \( x = \frac{1}{\sqrt{e}} \) into the original function:

\( f\left(\frac{1}{\sqrt{e}}\right) = \left(\frac{1}{\sqrt{e}}\right)^2 \log\left(\frac{1}{\sqrt{e}}\right) = \frac{1}{e} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2e} \).

The correct answer is \( \frac{-1}{2e} \).