Question

Find the point (x,y) on the graph of y=rootx that is closest to the point (4,0). Please help!

Answer 1

The equation is \[\Large y = \sqrt{x}\] right?

Answer 2

yes

Answer 3

Any point (x,y) on this graph is going to be the point \(\Large \left(x, \sqrt{x}\right)\). Agreed? or no?

Answer 4

okay, got it

Answer 5

so here is how I would approach it

Answer 6

First point (fixed)
\[\Large P = (x_1,y_1) = (4,0)\]
\[\Large x_1 = 4\]
\[\Large y_1 = 0\]


Second point (moving)
\[\Large Q = (x_1,y_1) = \left(x, \sqrt{x}\right)\]
\[\Large x_1 = x\]
\[\Large y_1 = \sqrt{x}\]
where x is nonnegative


Distance from point P to point Q
(using the distance formula)
\[\Large d(P,Q) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]
\[\Large d(P,Q) = \sqrt{(4-x)^2+(0-\sqrt{x})^2}\]
\[\Large d(P,Q) = \sqrt{(4-x)^2+(-\sqrt{x})^2}\]
\[\Large d(P,Q) = \sqrt{16-8x+x^2+x}\]
\[\Large d(P,Q) = \sqrt{x^2-7x+16}\]


The function \[\Large d(x) = \sqrt{x^2-7x+16}\] represents the distance from point P (4,0) to point Q (any point on \(\Large y = \sqrt{x}\))


The goal is to minimize d(x)

Answer 7

Let me know if you're able to minimize d(x)

Answer 8

you could also try minimizing \(d^2\),
the extrema of \(|d|\) and \(|d|^2\) occur at the same \(x\) value

Answer 9

okay, would it be x^2-7/2rootx^2-7x+16?

Answer 10

\[\Large d(x) = \sqrt{x^2-7x+16}\]


\[\Large d \ ' (x) = \frac{2x-7}{2\sqrt{x^2-7x+16}}\]

Answer 11

you should have 2x-7 up top (not x^2-7)

Answer 12

Now solve \[\Large d \ ' (x) = 0\]

Answer 13

would my point be (7/2,rott7/2)?

Answer 14

x = 7/2 is correct
idk how you got \(\Large \sqrt{\frac{7}{2}}\) for the distance

Answer 15

oh nvm I'm not thinking

Answer 16

yes, the point \[\Large \left(\frac{7}{2}, \sqrt{\frac{7}{2}}\right)\] is the right final answer

Answer 17

okay! thank you!