For the function $f(x) = x^x, x > 0$, which of the following are TRUE?

(A) $f'(x) = x^x(1 + \log x)$

(B) $x = e$ is the critical point

(C) $f$ is increasing in $(\frac{1}{e}, \infty)$

(D) $f$ is increasing in $(0, \infty)$

Choose the correct answer from the options given below:

The function is $f(x) = x^x$ for $x > 0$.

To find $f'(x)$, use logarithmic differentiation.

Let $y = x^x$

$\ln(y) = \ln(x^x)$

$\ln(y) = x \ln(x)$

Differentiate both sides with respect to $x$:

$\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x}$

$\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + 1$

$\frac{dy}{dx} = y(\ln(x) + 1)$

$f'(x) = x^x(1 + \ln(x))$

Statement (A) is TRUE.

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Critical points occur where $f'(x) = 0$:

$x^x(1 + \ln(x)) = 0$

Since $x^x > 0$ for all $x > 0$:

$1 + \ln(x) = 0$

$\ln(x) = -1$

$x = e^{-1}$

$x = \frac{1}{e}$

The critical point is $x = \frac{1}{e}$, not $x = e$.

Statement (B) is FALSE.

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The function is increasing where $f'(x) > 0$:

$x^x(1 + \ln(x)) > 0$

Since $x^x > 0$ for all $x > 0$:

$1 + \ln(x) > 0$

$\ln(x) > -1$

$x > \frac{1}{e}$

For $x \in (0, \frac{1}{e})$: $f'(x) < 0$, so $f$ is decreasing.

For $x \in (\frac{1}{e}, \infty)$: $f'(x) > 0$, so $f$ is increasing.

Statement (C) is TRUE.

Statement (D) is FALSE.

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Statements (A) and (C) are TRUE.