University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.5 - Exponential Functions - Exercises - Page 38: 23


The domain of $g(t)$ is $(-\infty,\infty)$ The range of $g(t)$ is $(1,\infty)$

Work Step by Step

$$g(t)=\sqrt{1+3^{-t}}=\sqrt{1+\frac{1}{3^t}}$$ 1) Domain: $g(t)$ is only defined as it satisfies 2 condititons: $3^t\ne0$ and $(1+3^{-t})\ge0$ - First, for $3^t\ne0$: since $3\gt0$, we have $3^t\gt0$ for all $t\in R$. Therefore, there is no value of $t$ which would lead to $3^t=0$. - Second, for $(1+3^{-t})\ge0$: As $3^t\gt0$ for all $t\in R$, $\frac{1}{3^t}=3^{-t}\gt0$ for all $t\in R$. That means $1+3^{-t}\gt1$ for all $t\in R$ Therefore, g(t) is defined on $R$. In other words, the domain of $g(t)$ is $(-\infty,\infty)$ 2) Range: - As proved above, $1+3^{-t}\gt1$ as $t$ in the domain. So $\sqrt{1+3^{-t}}\gt1$ In other words, the range of g(t) is $[-1,1]$ - Second, since $t$ is defined up until $-\infty$, $3^t$ can reach value as low as $0$ (though never equals $0$), meaning $\sqrt{1+\frac{1}{3^t}}$ can range up to $\infty$. Therefore, the range of $g(t)$ would be $(1,\infty)$.
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