#### Answer

$f(x)$ is continuous for all values of x and has no discontinuities.

#### Work Step by Step

$x^2+1=0\to x^2=-1$ which has no real solution.
The denominator can never be zero, and the function is defined and continuous for all x.

Published by
Brooks Cole

ISBN 10:
1-28505-709-0

ISBN 13:
978-1-28505-709-5

$f(x)$ is continuous for all values of x and has no discontinuities.

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