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Ch. 2 - Graphs and Functions
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. 2 - Graphs and FunctionsProblem 83b
Chapter 3, Problem 83b

Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7.
ƒ(x)=√(x+2),g(x)=−(1x)ƒ(x)=\(\surd\)(x+2),g(x)=-(\(\frac{1}{x}\))

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Step 1: Understand the composition of functions. For (ƒ∘g)(x), this means ƒ(g(x)), which is the function ƒ applied to the output of g(x). Similarly, for (g∘ƒ)(x), this means g(ƒ(x)).
Step 2: Find (ƒ∘g)(x) by substituting g(x) into ƒ. Since ƒ(x) = \(\sqrt{x+2}\), replace x with g(x) = -\(\frac{1}{x}\) to get (ƒ∘g)(x) = \(\sqrt{-\frac{1}{x}\) + 2}.
Step 3: Determine the domain of (ƒ∘g)(x). The expression inside the square root must be greater than or equal to zero: \(\left\)(-\(\frac{1}{x}\) + 2\(\right\)) \(\geq\) 0. Also, consider the domain restrictions of g(x), which excludes x = 0 because of division by zero.
Step 4: Find (g∘ƒ)(x) by substituting ƒ(x) into g. Since g(x) = -\(\frac{1}{x}\), replace x with ƒ(x) = \(\sqrt{x+2}\) to get (g∘ƒ)(x) = -\(\frac{1}{\sqrt{x+2}\)}.
Step 5: Determine the domain of (g∘ƒ)(x). The denominator \(\sqrt{x+2}\) cannot be zero, so x + 2 > 0, which means x > -2. Also, since the square root is in the denominator, x = -2 is excluded to avoid division by zero.

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Key Concepts

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

Function Composition

Function composition involves applying one function to the result of another, denoted as (f∘g)(x) = f(g(x)). It requires substituting the entire output of g(x) into f(x), creating a new function that combines both operations.
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Function Composition

Domain of a Function

The domain is the set of all input values for which a function is defined. When composing functions, the domain of the composite function depends on the domain of the inner function and the domain restrictions of the outer function after substitution.
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Domain Restrictions of Composed Functions

Square Root and Rational Function Restrictions

The square root function requires the radicand to be non-negative, limiting inputs to values where x+2 ≥ 0. Rational functions like g(x) = -1/x are undefined at x = 0, so these restrictions must be considered when determining the domain of compositions.
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Restrictions on Rational Equations
Related Practice
Textbook Question

Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7.

ƒ(x)=√x,g(x)=1(x+5)ƒ(x)=\(\surd\) x,g(x)=\(\frac{1}{(x+5)}\)

Textbook Question

Graph each function. See Examples 6–8 and the Summary of Graphing Techniques box following Example 9. ƒ(x)=2√x+1

Textbook Question

Given functions f and g, find (a)(ƒ∘g)(x) and its domain, and (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7.

ƒ(x)=4x,g(x)=x+4ƒ(x)=\(\frac{4}{x}\),g(x)=x+4

Textbook Question

Given functions f and g, find (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7.

ƒ(x)=2x,g(x)=x+1ƒ(x)=\(\frac{2}{x}\),g(x)=x+1

Textbook Question

Given functions f and g, find (a)(ƒ∘g)(x) and its domain, and (b)(g∘ƒ)(x) and its domain. See Examples 6 and 7.

ƒ(x)=√(x+4),g(x)=−(2x)ƒ(x)=\(\surd\)(x+4),g(x)=-(\(\frac{2}{x}\))

Textbook Question

Given functions f and g, find (a)(ƒ∘g)(x) and its domain. See Examples 6 and 7.

ƒ(x)=√(x+2),g(x)=−(1x)ƒ(x)=\(\surd\)(x+2),g(x)=-(\(\frac{1}{x}\))

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