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Q59.

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Precalculus Mathematics for Calculus

Found in: Page 157

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

Find the domain of the function $f (t) = \sqrt{t + 1}$ .

The domain $(D_{f})$ of the function is $t \in [- 1, \infty)$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. By definition of domain of a function.

The given function is $f (t) = \sqrt{t + 1}$ . Now, by definition of domain of function, the domain of the function is defined for all real values of $(t)$ for which the function is defined.

Step 2. By definition of domain of square root function.

From the square root function $f (t) = \sqrt{t + 1}$ , the domain is defined for all real numbers $(t)$ for which the expression inside the square root is greater than or equal to zero.

role="math" localid="1644580287740" $\begin{matrix} t + 1 \geq 0 gives \\ t \geq - 1 \end{matrix}$

Hence the domain of the function is $t \in [- 1, \infty)$

Step 3. Description of steps.

The domain of the function is $t \in [- 1, \infty)$ .

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