Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 60

Answer

All real numbers.

Work Step by Step

The domain of $g(t)=\sqrt {t^2+9}$ is the set of all real numbers $t$ for which $\sqrt{t^2+9}$ is defined as a real number. We know the sqare root of a negative number is not defined, so the domain of $g(t)=\sqrt{t^2+9}$ is all real numbers $t$ for which $t^2+9 \geq 0$. We know $t^2 \geq 0$ for all real numbers $t,$ and, therefore, $$t^2 + 9 \geq t^2 \geq 0$$ for all real nmbers $t$, which gives $$t^2 +9 \geq 0$$ for all real numbers $t$. Thus the domain of this function is all real numbers. In set notation, the domain is $\{t:t\in R\}.$ In interval notation, the domain is $(-\infty , \infty).$
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