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We are given the functions \( f(x) = \frac{1 - x}{22x} \) and \( g(x) = \frac{1}{1 + 22x} \), and asked to find the compositions \( (f \circ g)(x) \) and \( (g \circ f)(x) \), along with their domains.

### 1. Find \( (f \circ g)(x) \)

This means we substitute \( g(x) \) into \( f(x) \):

\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{1 + 22x}\right)
\]

Now substitute \( g(x) \) into the expression for \( f(x) \):

\[
f\left( \frac{1}{1 + 22x} \right) = \frac{1 - \frac{1}{1 + 22x}}{22 \cdot \frac{1}{1 + 22x}}
\]

Simplify the expression:

\[
1 - \frac{1}{1 + 22x} = \frac{(1 + 22x) - 1}{1 + 22x} = \frac{22x}{1 + 22x}
\]
\[
22 \cdot \frac{1}{1 + 22x} = \frac{22}{1 + 22x}
\]

Thus,

\[
f\left( \frac{1}{1 + 22x} \right) = \frac{\frac{22x}{1 + 22x}}{\frac{22}{1 + 22x}} = x
\]

So, \( (f \circ g)(x) = x \).

### 2. Find \( (g \circ f)(x) \)

This means we substitute \( f(x) \) into \( g(x) \):

\[
(g \circ f)(x) = g(f(x)) = g\left( \frac{1 - x}{22x} \right)
\]

Now substitute \( f(x) \) into the expression for \( g(x) \):

\[
g\left( \frac{1 - x}{22x} \right) = \frac{1}{1 + 22 \cdot \frac{1 - x}{22x}}
\]

Simplify the denominator:

\[
1 + 22 \cdot \frac{1 - x}{22x} = 1 + \frac{1 - x}{x} = \frac{x + (1 - x)}{x} = \frac{1}{x}
\]

Thus,

\[
g\left( \frac{1 - x}{22x} \right) = \frac{1}{\frac{1}{x}} = x
\]

So, \( (g \circ f)(x) = x \).

### 3. Domains of \( f \circ g \) and \( g \circ f \)

- For \( (f \circ g)(x) \), since both \( f(x) \) and \( g(x) \) simplify to \( x \), the domain will be all values of \( x \) where \( g(x) \) is defined. The domain of \( g(x) \) is all \( x \) such that \( 1 + 22x \neq 0 \), or \( x \neq -\frac{1}{22} \). So, the domain of \( f \circ g \) is \( (-\infty, -\frac{1}{22}) \cup (-\frac{1}{22}, \infty) \).

- For \( (g \circ f)(x) \), similarly, we need \( f(x) \) to be defined, which requires \( 22x \neq 0 \), or \( x \neq 0 \). So, the domain of \( g \circ f \) is \( (-\infty, 0) \cup (0, \infty) \).

### Summary:

- \( (f \circ g)(x) = x \) with domain \( (-\infty, -\frac{1}{22}) \cup (-\frac{1}{22}, \infty) \)
- \( (g \circ f)(x) = x \) with domain \( (-\infty, 0) \cup (0, \infty) \)

Would you like more details or have any questions?

Here are some related questions for you:
1. How do you find the composition of two functions?
2. What is the domain of a function and how do you determine it?
3. Why does \( f \circ g \) simplify to \( x \) in this case?
4. What are some cases where function compositions do not result in simple identities?
5. Can the domains of composed functions differ from the domains of the individual functions?

**Tip:** When finding domains of composite functions, always check the restrictions from both functions in the composition!

For the given functions, find (f ∘ g)(x) and (g ∘ f)(x) and the domain of each. f(x) = (1 - x) / 22x, g(x) = 1 / (1 + 22x)

Learn how to find the compositions (f ∘ g)(x) and (g ∘ f)(x) and their domains for the functions f(x) = (1 - x) / 22x and g(x) = 1 / (1 + 22x). This problem covers function composition and finding the domain of the resulting functions. Suitable for high school students in Grades 10-12.

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