The minimum value of f(x) = xlog x , x gt 0 , is

The correct answer is Option (4) → \(-\displaystyle\frac{1}{e}\)

By the product rule:

    \( \displaystyle f'(x) = \log x + x \cdot \frac{1}{x} = \log x + 1 \)

Setting f’(x) = 0 gives \(\log x = -1\), so the only critical point is \(x = \displaystyle\frac{1}{e}\).

    \( \displaystyle f''(x) = \frac{1}{x} \gt 0 \)

So f has its minimum at \(x = \frac{1}{e}\), and the minimum value is

    \( \displaystyle f\left(\frac{1}{e}\right) = \frac{1}{e}\log\frac{1}{e} = -\frac{1}{e} \)

Why the other options are incorrect

• Option (1) 0 — Incorrect. This is f(1); f takes smaller (negative) values for 0 < x < 1.

• Option (2) \(\displaystyle\frac{1}{e}\) — Incorrect. This is the point of minimum, not the minimum value.

• Option (3) \(-e\) — Incorrect. f is never this small; its least value is \(-\frac{1}{e} \approx -0.37\).