## Calculus: Early Transcendentals 8th Edition

$(\frac{5}{2},\sqrt{\frac{5}{2}})$ is the point on the curve $y = \sqrt{x}$ that is closest to the point $(3,0)$
All the points on the curve $y = \sqrt{x}$ have the form $(x, \sqrt{x})$ Note that $x \geq 0$ and $y \geq 0$ We can write an expression for the distance from points on the curve to the point $(3,0)$: $d = \sqrt{(x-3)^2+(y-0)^2}$ $d = \sqrt{(x-3)^2+(\sqrt{x})^2}$ $d = \sqrt{x^2-6x+9+x}$ $d = \sqrt{x^2-5x+9}$ We can find the point where $d'(x) = 0$: $d(x) = \sqrt{x^2-5x+9}$ $d'(x) = \frac{2x-5}{2\sqrt{x^2-5x+9}} = 0$ $2x-5 = 0$ $x = \frac{5}{2}$ When $0 \leq x \lt \frac{5}{2}$, then $d'(x) \lt 0$ When $x \gt \frac{5}{2}$, then $d'(x) \gt 0$ Thus, $x=\frac{5}{2}$ is the point where $d(x)$ is a minimum. We can find $y$: $y = \sqrt{x} = \sqrt{\frac{5}{2}}$ $(\frac{5}{2},\sqrt{\frac{5}{2}})$ is the point on the curve $y = \sqrt{x}$ that is closest to the point $(3,0)$