Question

If $f(x) = x^x$, $x > 0$, then

  • $f(x)$ is increasing on $(1/e, \infty)$
  • $f(x)$ is decreasing on $(0, 1/e)$
  • $x = 1/e^{2}$ is the point of local minima
  • The local minimum value of $f(x)$ is $e^{-1/e}$.

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Answer


The correct options are:

  • A: $f(x)$ is increasing on $( \frac{1}{e}, \infty )$
  • B: $f(x)$ is decreasing on $( 0, \frac{1}{e} )$
  • D: The local minimum value of $f(x) = \frac{1}{e}$ is $e^{-\frac{1}{e}}$

To understand why these options are correct, let's analyze the function $f(x) = x^x$ for $x > 0$.

First, compute the derivative of $f(x)$: $$ f'(x) = x^x (1 + \log x) $$

For $f'(x) > 0$: $$ x^x (1 + \log x) > 0 $$ This implies: $$ 1 + \log x > 0 \Rightarrow x > \frac{1}{e} $$ Hence, $f(x)$ is increasing for $x \in \left( \frac{1}{e}, \infty \right)$.

For $f'(x) < 0$: $$ x^x (1 + \log x) < 0 $$ This implies: $$ 1 + \log x < 0 \Rightarrow x < \frac{1}{e} $$ Therefore, $f(x)$ is decreasing for $x \in \left( 0, \frac{1}{e} \right)$.

For finding the local minimum, set $f'(x) = 0$: $$ 1 + \log x = 0 \Rightarrow x = \frac{1}{e} $$

Evaluate $f$ at $x = \frac{1}{e}$: $$ f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{\frac{1}{e}} = e^{-\frac{1}{e}} $$

Thus, the local minimum value of $f(x)$ is $e^{-\frac{1}{e}}$.


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