## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 10: 58

#### Answer

$f$ has no zeros.

#### Work Step by Step

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.

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