#### Answer

Slope of Tangent Line: $\frac{1}{5}$

#### Work Step by Step

$f(x) = \tan^{-1}(x)$
$f'(x) = \frac{1}{1+x^2}$
$f'(-2)=\frac{1}{1+(-2)^2} = \frac{1}{5}$

Published by
Pearson

ISBN 10:
0321947347

ISBN 13:
978-0-32194-734-5

Slope of Tangent Line: $\frac{1}{5}$

$f(x) = \tan^{-1}(x)$
$f'(x) = \frac{1}{1+x^2}$
$f'(-2)=\frac{1}{1+(-2)^2} = \frac{1}{5}$

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