For the function $f(x) = x^{1/x}$, $x > 0$, which of the following are correct?

(A) $x = 0$ is the only point where extremum may occur.

(B) The given function is maximum at $x = e$.

(C) The function has no extreme value for $x > 0$.

(D) The maximum value of the function $f(x)$ is $e^{1/e}$.

Choose the correct answer from the options given below:

Given the function $f(x) = x^{1/x}$ for $x > 0$.

Taking natural logarithm on both sides:

$\ln f(x) = \ln(x^{1/x})$

$\ln f(x) = \frac{1}{x} \cdot \ln x$

$\ln f(x) = \frac{\ln x}{x}$

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Differentiating both sides with respect to $x$:

$\frac{f'(x)}{f(x)} = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2}$

$\frac{f'(x)}{f(x)} = \frac{1 - \ln x}{x^2}$

$f'(x) = x^{1/x} \cdot \frac{1 - \ln x}{x^2}$

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Setting $f'(x) = 0$:

$x^{1/x} \cdot \frac{1 - \ln x}{x^2} = 0$

Since $x^{1/x} > 0$ and $x^2 > 0$ for all $x > 0$:

$1 - \ln x = 0$

$\ln x = 1$

$x = e$

The only critical point is $x = e$.

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For $x < e$: $\ln x < 1$, so $1 - \ln x > 0$, thus $f'(x) > 0$ (function is increasing)

For $x > e$: $\ln x > 1$, so $1 - \ln x < 0$, thus $f'(x) < 0$ (function is decreasing)

Therefore, $x = e$ is a maximum point.

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The maximum value:

$f(e) = e^{1/e}$

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(A) $x = 0$ is not in the domain. The extremum occurs at $x = e$. Incorrect.

(B) The function is maximum at $x = e$. Correct.

(C) The function has a maximum at $x = e$. Incorrect.

(D) The maximum value is $e^{1/e}$. Correct.

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Statements (B) and (D) are correct.