Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 47E
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Suppose that the functions u and w are defined as follows
![**Composite Functions Example**
Suppose that the functions \( u \) and \( w \) are defined as follows:
\[ u(x) = x^2 + 6 \]
\[ w(x) = \sqrt{x + 9} \]
Find the following composite functions:
1. \( (u \circ w)(7) \)
2. \( (w \circ u)(x) \)
**Graphical Explanation:**
To understand how to evaluate these composite functions, follow these steps:
- Determine the inner function's result first.
- Use this result as the input for the outer function.
### For \( (u \circ w)(7) \):
Evaluate \( w(7) \):
\[ w(7) = \sqrt{7 + 9} = \sqrt{16} = 4 \]
Now use \( w(7) = 4 \) as the input for \( u(x) \):
\[ u(4) = 4^2 + 6 = 16 + 6 = 22 \]
So, \( (u \circ w)(7) = 22 \).
### For \( (w \circ u)(x) \):
Evaluate \( u(x) \):
\[ u(x) = x^2 + 6 \]
Now use \( u(x) \) as the input for \( w(x) \):
\[ w(u(x)) = w(x^2 + 6) = \sqrt{x^2 + 6 + 9} = \sqrt{x^2 + 15} \]
So, \( (w \circ u)(x) = \sqrt{x^2 + 15} \).
### Summary:
- \( (u \circ w)(7) = 22 \)
- \( (w \circ u)(x) = \sqrt{x^2 + 15} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64f54e3f-bbb2-46db-a1a9-bb995302d054%2F88181d75-5a75-4f34-8967-b45802ff5f46%2F34ehwwt.jpeg&w=3840&q=75)
Transcribed Image Text:**Composite Functions Example**
Suppose that the functions \( u \) and \( w \) are defined as follows:
\[ u(x) = x^2 + 6 \]
\[ w(x) = \sqrt{x + 9} \]
Find the following composite functions:
1. \( (u \circ w)(7) \)
2. \( (w \circ u)(x) \)
**Graphical Explanation:**
To understand how to evaluate these composite functions, follow these steps:
- Determine the inner function's result first.
- Use this result as the input for the outer function.
### For \( (u \circ w)(7) \):
Evaluate \( w(7) \):
\[ w(7) = \sqrt{7 + 9} = \sqrt{16} = 4 \]
Now use \( w(7) = 4 \) as the input for \( u(x) \):
\[ u(4) = 4^2 + 6 = 16 + 6 = 22 \]
So, \( (u \circ w)(7) = 22 \).
### For \( (w \circ u)(x) \):
Evaluate \( u(x) \):
\[ u(x) = x^2 + 6 \]
Now use \( u(x) \) as the input for \( w(x) \):
\[ w(u(x)) = w(x^2 + 6) = \sqrt{x^2 + 6 + 9} = \sqrt{x^2 + 15} \]
So, \( (w \circ u)(x) = \sqrt{x^2 + 15} \).
### Summary:
- \( (u \circ w)(7) = 22 \)
- \( (w \circ u)(x) = \sqrt{x^2 + 15} \)
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