Represent any point on the curve by (x, 1-x^2). The distance between (0, 0) and (x, 1-x^2) is
[tex] \sqrt{(x-0)^2+(1-x^2-0)^2}=\sqrt{x^2+(1-x^2)^2}=\sqrt{x^2+1-2x^2+x^4} [/tex]
To make this easier, let's minimize the SQUARE of this quantity because when the square root is minimal, its square will be minimal.
So minimize [tex]L=x^4-x^2+1[/tex]
Find the derivative of L and set it equal to zero.
[tex]\frac{d}{dx}(L)=4x^3-2x \\ 4x^3-2x=0 \\ 2x(2x^2-1)=0[/tex]
This gives you [tex]x=0[/tex] or [tex]x^2=\frac{1}{2} \\ x=\pm\sqrt{2}/2[/tex]
You can use the Second Derivative Test to figure out which value(s) produce the MINIMUM distance.
[tex]\frac{d^2}{dx}=12x^2-2[/tex]
When x = 0, the second derivative is negative, indicating a relative maximum. When [tex]x=\pm\frac{\sqrt{2}}{2}[/tex], the second derivative is positive, indicating a relative MINIMUM.
The two points on the curve closest to the origin are [tex]\left( \pm\frac{\sqrt{2}}{2},\frac{1}{2} \right)[/tex]
