Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises - Page 10: 38


$g(x)=\frac{\sqrt{2}}{x^6+x^2+1}$ and $f(x)=x^2$ The domain of $h$ is $\mathbb{R}$.

Work Step by Step

If we set $g(x)=\frac{\sqrt{2}}{x^6+x^2+1}$ and $f(x)=x^2$, then: $f \circ g=f(g(x))=(\frac{\sqrt{2}}{x^6+x^2+1})^2=\frac{2}{(x^6+x^2+1)^2}$ The domain of $h$ is $\mathbb{R}$ (because the denominator is never 0).
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