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Part 1
For f left parenthesis x right parenthesis equals StartFraction 9 Over x plus 6 EndFraction
 and g left parenthesis x right parenthesis equals StartFraction 1 Over x EndFraction
​, find
a.  left parenthesis f circle g right parenthesis left parenthesis x right parenthesis​;     b.  the domain of f circle g
Question content area bottom
Part 1
a.  left parenthesis f circle g right parenthesis left parenthesis x right parenthesisequals
  
StartFraction 9 x Over 1 plus 6 x EndFraction
​(Simplify your​ answer.)
Part 2
b.  What is the domain of f circle g​?
The domain is 
  
left parenthesis negative infinity comma negative one sixth right parenthesis union left parenthesis negative one sixth comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
.  
​(Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the​ expression.)

We are given two functions \( f(x) = \frac{9}{x+6} \) and \( g(x) = \frac{1}{x} \). We need to find:

1. \( (f \circ g)(x) \) (which means \( f(g(x)) \)).
2. The domain of \( f \circ g \).

### Part 1: Find \( (f \circ g)(x) \)
To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \):

\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)
\]

Now, substitute \( \frac{1}{x} \) for \( x \) in \( f(x) \):

\[
f\left(\frac{1}{x}\right) = \frac{9}{\frac{1}{x} + 6}
\]

Simplify the expression:

\[
f\left(\frac{1}{x}\right) = \frac{9}{\frac{1}{x} + 6} = \frac{9}{\frac{1 + 6x}{x}} = \frac{9x}{1 + 6x}
\]

So,

\[
(f \circ g)(x) = \frac{9x}{1 + 6x}
\]

### Part 2: Domain of \( f \circ g \)

To find the domain of \( f \circ g \), we need to consider the restrictions from both \( f(x) \) and \( g(x) \).

- For \( g(x) = \frac{1}{x} \), the function is undefined when \( x = 0 \) (since division by zero is not allowed). Therefore, \( x \neq 0 \).
- For \( f(x) = \frac{9}{x+6} \), the function is undefined when \( x+6 = 0 \), or \( x = -6 \).

Now, since \( f(g(x)) = \frac{9x}{1 + 6x} \), we have to ensure that the denominator \( 1 + 6x \neq 0 \), which gives \( x \neq -\frac{1}{6} \).

Thus, the domain excludes \( x = 0 \) and \( x = -\frac{1}{6} \).

Therefore, the domain of \( f \circ g \) is all real numbers except 0 and \( -\frac{1}{6} \), written in interval notation as:

\[
\text{Domain of } f \circ g: (-\infty, -\frac{1}{6}) \cup (-\frac{1}{6}, 0) \cup (0, \infty)
\]

---

Would you like a more detailed explanation of any part? Here are 5 related questions for practice:

1. What is the domain of \( f(x) = \frac{9}{x+6} \)?
2. Find \( (g \circ f)(x) \) for \( f(x) = \frac{9}{x+6} \) and \( g(x) = \frac{1}{x} \).
3. How do you determine the domain of a composition of two functions?
4. Can a function have more than one domain restriction?
5. Why is \( x = 0 \) excluded from the domain of \( g(x) = \frac{1}{x} \)?

**Tip**: Always simplify the composition of functions before determining the domain to avoid unnecessary restrictions.

For f(x) = 9/(x + 6) and g(x) = 1/x, find (f ∘ g)(x) and the domain of f ∘ g.

Solve the composition of functions problem involving f(x) = 9/(x + 6) and g(x) = 1/x. Learn how to find (f ∘ g)(x) and determine its domain, highlighting key concepts in function composition and domain restrictions. Ideal for high school students in Grades 10-12.

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