Find the point on the graph of $$F(x)=1-x^{2}$$,those are closest to (0, 0).

Question

anonymous · Answer

the curve is following: http://www.wolframalpha.com/input/?i=y%3D+1+-+x%5E2

anonymous · Answer

I am not asking the curve lol.

hartnn · Answer

y^2+x^2 = r^2
minimize r.
(1-x^2)^2 +x^2 = r^2
r^2 =1+x^4 + -x^2
diff, r  and set it to 0
is this the approach ?

anonymous · Answer

Correct, keep going and type the answer.

hartnn · Answer

4x^2=2
x=+/-  sqrt(1/2)
y= 1/2
points (1/2,1/2),(-1/2,1/2)

anonymous · Answer

Great it is .

anonymous · Answer

Insert sqrt in the 1/2 and -1/2 of the  X coordinates.

hartnn · Answer

oh, yes. typo.

calculusfunctions · Answer

You want the minimum (shortest) distance from the given point to the parabola.

Let (x, y) be any point on the parabola.

Using the distance formula, we obtain

$$L =\sqrt{x ^{2}+y ^{2}}$$

Now substitute y = 1 - x² to get

$$L =\sqrt{x ^{2}+(1-x ^{2})^{2}}$$

Squaring both sides yields

$$L ^{2}=x ^{2}+(1-x ^{2})^{2}$$

Differentiating both sides w.r.t. x yields

$$2L \frac{ dL }{ dx }=2x +2(1-x ^{2})(-2x)$$

For max or min, dL/dx = 0. Hence

0 = 2x - 4x(1 - x²)

Factoring 2x from the right side yields

0 = 2x[1 - 2(1 - x²)]

0 = 2x(1 - 2 + 2x²)

0 = 2x(2x² - 1)

$$x =0, x =\pm \frac{ \sqrt{2} }{ 2 }$$

Thus

$$F(\pm \frac{ \sqrt{2} }{ 2 })=\frac{ 1 }{ 2 }$$

F(0) = 1

∴ the points on the parabola closest to the origin are (-√2/2, 1/2), (0, 1) and 
(√2/2, 1/2).

anonymous · Answer

Great work, but (0, 1) is not the possible answer.

calculusfunctions · Answer

@Zekarias  Yes you're right! I forgot to perform the first or second derivative test to determine which critical points are minimum. Sorry!

anonymous · Answer

Thanks for your time. I wish I gave you the medal.

calculusfunctions · Answer

That's OK! Thanks anyway.

hartnn · Answer

i don't mind if u give him......

anonymous · Answer

hey, you can give them all,,,, lol

calculusfunctions · Answer

Thanks guys! lol