1. The problem is to understand and describe the graph of the function $y = x^x$. 2. The function $y = x^x$ is defined for $x > 0$ because for real numbers, raising a negative number to a real power can be undefined or complex. 3. The function can be rewritten using exponentials and logarithms as: $$y = x^x = e^{x \ln x}$$ 4. To analyze the graph, consider the domain $x > 0$. 5. At $x=0$, the function is not defined, but the limit as $x \to 0^+$ is: $$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln x} = e^0 = 1$$ 6. At $x=1$, $y = 1^1 = 1$. 7. To find extrema, differentiate: $$y' = \frac{d}{dx} e^{x \ln x} = e^{x \ln x} \cdot \frac{d}{dx} (x \ln x) = x^x (\ln x + 1)$$ 8. Set derivative to zero for critical points: $$x^x (\ln x + 1) = 0 \implies \ln x = -1 \implies x = e^{-1} = \frac{1}{e} \approx 0.3679$$ 9. At $x = \frac{1}{e}$, the function has a minimum. 10. The minimum value is: $$y\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{\frac{1}{e}} = e^{-\frac{1}{e}} \approx 0.6922$$ 11. For $x > 1$, the function increases rapidly. 12. Summary: The graph of $y = x^x$ is defined for $x > 0$, approaches 1 as $x \to 0^+$, has a minimum at $x = \frac{1}{e}$ with value approximately 0.6922, and increases rapidly for $x > 1$. Final answer: The graph of $y = x^x$ is an increasing curve for $x > 0$ with a minimum at $x = \frac{1}{e}$.