## Calculus: Early Transcendentals (2nd Edition)

$\frac{1}{y^{6}+1}$
$f(x) = \frac{1}{x^{3}+1}$ Evaule $f(2)$ $f(2)= \frac{1}{2^{3}+1} = \frac{1}{9}$ Evaluate $f(y^{2})$ $f(y^{2})=\frac{1}{(y^{2})^{3}+1} = \frac{1}{y^{6}+1}$