Question

Find the points on the graph of y=4-x^2 that are closest to the point (0,2)

Answer 1

@agent0smith

Answer 2

I think you'll have to use the distance formula. let x2 be x, x1 = 0, y2 = 4-x^2, y1 = 2
\[d ^{2}=(x-0)^{2}+((4-x ^{2})-2)^{2}\]

Answer 3

k

Answer 4

And then you want to minimize that, so differentiate and set it to zero

Answer 5

\[d=\sqrt{(x-0)^2 + (y-2)^2}\]

Answer 6

\[d=\sqrt{x^2+(4-x^2-2)^2}\]

Answer 7

Just leave it as d^2, since if you minimize d^2, you also minimize d. Also the square root will disappear later anyway, even if you did use it.

Answer 8

\[d=\sqrt{x^2+(2-x^2)^2}\]

Answer 9

\[d'=\frac{ 1 }{ 2 } \times (x^2+(2-x^2)^2)^\frac{ -1 }{ 2 } \times 2x +2(2-x^2)(-2x)\]

Answer 10

\[\frac{ 2x-4x(2-x^2) }{ 2\sqrt{x^2+(2-x^2)^2)} } = 0\]