Curve Points 1D2Aea

1. **State the problem:** We are given the curve $$y = x^{1/2} - 4 + 3x^{-1/2}$$ and need to find: - The coordinates of points $A$ and $B$ where the curve crosses the $x$-axis. - The coordinates of the minimum point $C$, with $y$ in the form $a\sqrt{3} + b$. 2. **Find points $A$ and $B$ (where $y=0$):** Set $$y = 0 = x^{1/2} - 4 + 3x^{-1/2}$$ Rewrite $x^{1/2} = \sqrt{x}$ and $x^{-1/2} = \frac{1}{\sqrt{x}}$: $$0 = \sqrt{x} - 4 + \frac{3}{\sqrt{x}}$$ Multiply both sides by $\sqrt{x}$ (assuming $x>0$): $$0 = x - 4\sqrt{x} + 3$$ Let $t = \sqrt{x}$, then: $$0 = t^2 - 4t + 3$$ Factor the quadratic: $$0 = (t - 3)(t - 1)$$ So $t = 1$ or $t = 3$. Recall $t = \sqrt{x}$, so: $$\sqrt{x} = 1 \implies x = 1$$ $$\sqrt{x} = 3 \implies x = 9$$ Find corresponding $y$ values (should be zero): At $x=1$: $$y = 1 - 4 + 3 = 0$$ At $x=9$: $$y = 3 - 4 + \frac{3}{3} = 3 - 4 + 1 = 0$$ Thus, points $A$ and $B$ are: $$A = (1, 0), \quad B = (9, 0)$$ 3. **Find the minimum point $C$:** To find minima, differentiate $y$ with respect to $x$: $$y = x^{1/2} - 4 + 3x^{-1/2}$$ $$\frac{dy}{dx} = \frac{1}{2}x^{-1/2} + 0 - \frac{3}{2}x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{3}{2x^{3/2}}$$ Set derivative to zero for critical points: $$0 = \frac{1}{2\sqrt{x}} - \frac{3}{2x^{3/2}}$$ Multiply both sides by $2x^{3/2}$: $$0 = x - 3$$ So: $$x = 3$$ Check second derivative to confirm minimum: $$\frac{d^2y}{dx^2} = -\frac{1}{4}x^{-3/2} + \frac{9}{4}x^{-5/2}$$ At $x=3$: $$\frac{d^2y}{dx^2} = -\frac{1}{4(3^{3/2})} + \frac{9}{4(3^{5/2})} > 0$$ So $x=3$ is a minimum. Find $y$ at $x=3$: $$y = \sqrt{3} - 4 + \frac{3}{\sqrt{3}} = \sqrt{3} - 4 + \sqrt{3} = 2\sqrt{3} - 4$$ Thus, minimum point $C$ is: $$C = \left(3, 2\sqrt{3} - 4\right)$$ **Final answers:** - $A = (1, 0)$ - $B = (9, 0)$ - $C = \left(3, 2\sqrt{3} - 4\right)$