To find the minimum value of the function $f(x) = x^x$, we can proceed by first rewriting it in a form that simplifies differentiation, allowing us to locate any critical points.

Step 1: Rewrite the Function Using Logarithms

Rewrite $f(x) = x^x$ in a way that allows differentiation: $f(x) = e^{x \ln x}$ Let $g(x) = x \ln x$, so that $f(x) = e^{g(x)}$. The minimum of $f(x)$ will occur where $g(x)$ reaches its minimum, since $e^x$ is an increasing function.

Step 2: Find the Derivative of $g(x)$

To find the critical points of $g(x) = x \ln x$, we take its derivative: $g'(x) = \ln x + 1$ Set $g'(x) = 0$ to find critical points: $\ln x + 1 = 0 \Rightarrow \ln x = -1 \Rightarrow x = e^{-1} = \frac{1}{e}$

Step 3: Determine the Nature of the Critical Point

To confirm that $x = \frac{1}{e}$ gives a minimum, consider the second derivative of $g(x)$: $g''(x) = \frac{1}{x}$ Since $g''(x) > 0$ for $x > 0$, the function $g(x)$ is concave up at $x = \frac{1}{e}$, indicating a minimum at this point.

Step 4: Calculate the Minimum Value

Now, evaluate $f(x)$ 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 minimum value of $f(x) = x^x$ is: $\boxed{e^{-\frac{1}{e}} \approx 0.6922}$

Would you like more details or have any questions?

Here are five related questions:

1. How can you find maximum points for similar functions?
2. What happens to $f(x) = x^x$ as $x \to \infty$ and $x \to 0^+$?
3. Can the minimum be obtained without calculus?
4. How does the derivative $g'(x) = \ln x + 1$ behave near $x = 1$?
5. Are there any other local extrema for $f(x) = x^x$?

Tip: Rewriting a function in terms of $e$ can simplify differentiation and reveal critical points easily.