1. The problem is to understand and analyze the function $F(x) = x - \sqrt{x}$. 2. The function involves a linear term $x$ and a square root term $\sqrt{x}$. The domain of $F(x)$ is $x \geq 0$ because the square root is defined only for non-negative $x$. 3. To find the intercepts, set $F(x) = 0$: $$x - \sqrt{x} = 0$$ 4. Factor the equation: $$\sqrt{x}(\sqrt{x} - 1) = 0$$ 5. Set each factor equal to zero: $$\sqrt{x} = 0 \implies x = 0$$ $$\sqrt{x} - 1 = 0 \implies \sqrt{x} = 1 \implies x = 1$$ 6. So the function crosses the x-axis at $x=0$ and $x=1$. 7. To find extrema, compute the derivative: $$F'(x) = 1 - \frac{1}{2\sqrt{x}}$$ 8. Set derivative equal to zero to find critical points: $$1 - \frac{1}{2\sqrt{x}} = 0 \implies \frac{1}{2\sqrt{x}} = 1 \implies 2\sqrt{x} = 1 \implies \sqrt{x} = \frac{1}{2} \implies x = \frac{1}{4}$$ 9. To determine if this is a minimum or maximum, check the second derivative: $$F''(x) = \frac{1}{4x^{3/2}} > 0 \text{ for } x > 0$$ Since $F''(\frac{1}{4}) > 0$, the function has a local minimum at $x=\frac{1}{4}$. 10. Evaluate $F(x)$ at $x=\frac{1}{4}$: $$F\left(\frac{1}{4}\right) = \frac{1}{4} - \sqrt{\frac{1}{4}} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$$ Final answer: The function $F(x) = x - \sqrt{x}$ has zeros at $x=0$ and $x=1$, and a local minimum at $x=\frac{1}{4}$ with value $-\frac{1}{4}$.