Question: Find the location of the points on $y=\frac 1x$ that are closest to the point $(-1,1)$.
Attempted answer: $f(x)=\sqrt{(x+1)^2+\left(\frac 1x-1\right)^2}$
taking the derivative of the function inside the square root gets me: $2(x+1)+(2\left(x^{-1}-1\right))\left(-x^{-2}\right)$
simplifying this gets me to: $2x+2-2x^{-3}+2x^{-2}$
so: $2(x+1-x^{-3}+x^{-2})=0$
I end up getting to: $x-x^{-3}+x^{-2}=-1$
I have no idea what to do at this point, although I feel like I'm probably doing something fundamentally wrong to begin with. I know the answer is supposed to end up at $(-1.6180,-0.6180),(0.6180,-1.680)$
Thanks for any help.
Your equation (simplified) can be factorized as $$2\, \left( {x}^{2}+1 \right) \left( {x}^{2}+x-1 \right) =0$$