To find the lowest value of $x^x$ where $x$ is a positive number, we need to evaluate the function $f(x) = x^x$ and find its minimum. Here's the step-by-step approach:

1. Rewrite the function in logarithmic form: $f(x) = x^x = e^{x \ln x}$ This allows us to differentiate the function using basic calculus rules.

2. Find the derivative: Differentiate $f(x) = e^{x \ln x}$ with respect to $x$. First, find the derivative of the exponent $x \ln x$ using the product rule: $\frac{d}{dx}(x \ln x) = \ln x + 1$ Therefore, the derivative of $f(x)$ is: $f'(x) = e^{x \ln x} \cdot (\ln x + 1) = x^x (\ln x + 1)$

3. Set the derivative to zero: To find the critical points where $f(x)$ has a minimum or maximum, set $f'(x) = 0$: $x^x (\ln x + 1) = 0$ Since $x^x \neq 0$ for any positive $x$, this simplifies to: $\ln x + 1 = 0$ Solving for $x$: $\ln x = -1 \quad \Rightarrow \quad x = e^{-1} = \frac{1}{e}$

4. Second derivative test: To confirm that $x = \frac{1}{e}$ is a minimum, we can compute the second derivative or analyze the behavior of $f(x)$ around this point. After performing the second derivative test, we find that this is indeed a minimum.

5. Calculate the minimum value: The minimum value of $f(x) = x^x$ occurs at $x = \frac{1}{e}$. The corresponding value of the function is: $f\left( \frac{1}{e} \right) = \left( \frac{1}{e} \right)^{1/e} = e^{-1/e}$ Numerically, this value is approximately $0.6922$.

Thus, the lowest value of $x^x$ for positive $x$ occurs at $x = \frac{1}{e}$, and the minimum value is $e^{-1/e} \approx 0.6922$.

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Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How can we apply this concept to find local minima in other exponential functions?
2. What happens to the behavior of $x^x$ as $x$ approaches 0 or infinity?
3. How does the minimum value change if we allow negative bases for $x$?
4. Can we generalize this process for $x^k$ where $k$ is a constant other than $x$?
5. What is the significance of $e$ in calculus and exponential functions?

Tip: When solving for extrema (minimum or maximum) in functions involving exponents, logarithmic differentiation can greatly simplify the problem.