How to find the set of points of to the curve $y^2=x^3+3x^2+x+1$ with the shortest distance to the point $(0,0)$
I try this:
$$d=\sqrt{x^2 + (x^3+3x^2+x+1)^2}$$ $$\frac{dd}{dx}= \frac{2x+2(x^3+3x^2+x+1)}{\sqrt{x^2 + (x^3+3x^2+x+1)^2}}=0$$
And now I would take the real solutions and substitute in $d$ to get the minimum distance from $(0,0)$ to the curve. But I'd have two problems; The only real solution is negative, and i would not be finding the closest points of the curve to my point, but only the minimum distance to one of them.
consider $$d^2=x^3+4x^2+x+1=g(x)$$ then we get $$g'(x)=3x^2+8x+1$$ and you must solve the equation $$g'(x)=0$$