Finding the closest point to the origin of $y=2\sqrt{\ln(x+3) }$ Given $$y=2\sqrt{\ln(x+3) }$$
How do I determine a (x,y) pair satisfying the above relation which is the closest to the origin (0,0)?

To minimize the distance from the origin, minimize the square of the distance from the origin, given by $d^2 = M = x^2+y^2$,
$$M = x^2+y^2=x^2+(2\sqrt{\ln(x+3)})^2 = x^2 + 4\ln(x+3)$$
$$\frac{dM}{dx} = 2x + \frac{4}{x+3} = \frac{2x(x+3) + 4}{x+3} = \frac{2x^2 + 6x + 4}{x+3}.$$
This is only (possibly) equal to zero when the numerator is equal to zero,
$$2x^2 + 6x + 4 = 0 \implies x=-2 \ \ \text{and} \ \ x=-1$$
We know that at least one of these must be where the minimum occurs, so just plug the values of $x$ into the given equation and compute the distance from the origin.
$y(-1) = 2\sqrt{\ln(-1+3)} = 2 \sqrt{\ln(2)} = 1.665 \implies d^2 = (-1)^2 + (1.665)^2 = 2.386$
$y(-2) = 2\sqrt{\ln(-2+3)} = 2 \sqrt{\ln(1)} = 0 \implies d^2 = (-1)^2 + (0)^2 = 1$
So the point nearest to the origin is $(-2,0)$.
