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Ch. 2 - Functions and Graphs
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 3, Problem 33a

Evaluate each function at the given values of the independent variable and simplify. f(r)=r+6+3f(r)=\(\sqrt{r + 6}\)+3
a. f(6)\(\text{a. }\)f(-6)

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Identify the given function: \(f(r) = \sqrt{r + 6} + 3\).
Substitute the given value \(r = -6\) into the function: \(f(-6) = \sqrt{-6 + 6} + 3\).
Simplify the expression inside the square root: \(-6 + 6 = 0\), so the expression becomes \(f(-6) = \sqrt{0} + 3\).
Evaluate the square root: \(\sqrt{0} = 0\), so the expression simplifies to \(f(-6) = 0 + 3\).
Add the terms to write the simplified form of \(f(-6)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Function Evaluation

Function evaluation involves substituting a given value for the independent variable into the function's expression and simplifying the result. For example, to find f(-6), replace every instance of the variable r with -6 and then simplify.
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Square Root Simplification

Simplifying square roots requires understanding that the square root of a number is a value that, when squared, gives the original number. When evaluating expressions like √(r + 6), ensure the expression inside the root is non-negative and simplify the root if possible.
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Domain of a Function

The domain is the set of all input values for which the function is defined. For functions involving square roots, the expression inside the root must be greater than or equal to zero to avoid imaginary numbers, so r + 6 ≥ 0 must hold true.
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Domain Restrictions of Composed Functions