## Calculus (3rd Edition)

$f$ has no zeros.
To find the zeros of $f(x)=\frac{1}{(x-1)^2+1}$, we set $$\frac{1}{(x-1)^2+1}=0$$ which has no solutions -- that is, the function $f$ has no zeros. Since $f(-x)\neq -f(x)$ nor $f(-x)\neq f(x)$ then the function is not symmetric about the origin nor about the y-axis and it is increasing on the interval $(-\infty,1)$. See the figure below.