Question

Find the poin on the graph 0f f(x)=1-x^2 that are closest to0(0,0)

Answer 1

Represent any point on the curve by (x, 1-x^2). The distance between (0, 0) and (x, 1-x^2) is

$\sqrt{(x-0)^2+(1-x^2-0)^2}=\sqrt{x^2+(1-x^2)^2}=\sqrt{x^2+1-2x^2+x^4}$

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 $L=x^4-x^2+1$

Find the derivative of L and set it equal to zero.

$\frac{d}{dx}(L)=4x^3-2x \\ 4x^3-2x=0 \\ 2x(2x^2-1)=0$

This gives you $x=0$ or $x^2=\frac{1}{2} \\ x=\pm\sqrt{2}/2$

You can use the Second Derivative Test to figure out which value(s) produce the MINIMUM distance.

$\frac{d^2}{dx}=12x^2-2$

When x = 0, the second derivative is negative, indicating a relative maximum.  When $x=\pm\frac{\sqrt{2}}{2}$, the second derivative is positive, indicating a relative MINIMUM.

The two points on the curve closest to the origin are $\left( \pm\frac{\sqrt{2}}{2},\frac{1}{2} \right)$