jmc

The question is asking us to find the smallest value of $x$ such that $x=2\cdot \frac{1}{x}-1$. We multiply through by $x$ to clear the fraction, then rearrange terms: $x^2+x-2=0$. This can be factored as $(x+2)(x-1)=0$. We could also use the quadratic formula to find $x$: $$x=\frac{-1\pm \sqrt{1^2-4(1)(-2)}}{2}.$$ Either way, we find that $x=1$ or $x=-2$. Since we want the smallest value of $x$, our answer is $\boxed{-2}$.