find f . g and g . f.                                                                                                                                                                                                    1.f(x)=3x-2; g(x) =1/3 (x+2)                                                                                                                                         2.f(x)=4x;g(x)=x^2-1                                                                                                                                                              3.f(x)=-x^2+1;g(x)=x                                                                                                                 

Let's solve each problem step by step by finding both \( f \circ g \) and \( g \circ f \), which represent the compositions of functions \( f(x) \) and \( g(x) \).

### 1. \( f(x) = 3x - 2 \) and \( g(x) = \frac{1}{3}(x + 2) \)

- **Find \( f \circ g \):**
  \[
  f(g(x)) = f\left(\frac{1}{3}(x + 2)\right)
  \]
  Substitute \( g(x) \) into \( f(x) \):
  \[
  f\left(\frac{1}{3}(x + 2)\right) = 3\left(\frac{1}{3}(x + 2)\right) - 2
  \]
  Simplify:
  \[
  f(g(x)) = (x + 2) - 2 = x
  \]
  So, \( f \circ g = x \).

- **Find \( g \circ f \):**
  \[
  g(f(x)) = g(3x - 2)
  \]
  Substitute \( f(x) \) into \( g(x) \):
  \[
  g(3x - 2) = \frac{1}{3}((3x - 2) + 2)
  \]
  Simplify:
  \[
  g(f(x)) = \frac{1}{3}(3x) = x
  \]
  So, \( g \circ f = x \).

### 2. \( f(x) = 4x \) and \( g(x) = x^2 - 1 \)

- **Find \( f \circ g \):**
  \[
  f(g(x)) = f(x^2 - 1)
  \]
  Substitute \( g(x) \) into \( f(x) \):
  \[
  f(x^2 - 1) = 4(x^2 - 1)
  \]
  Simplify:
  \[
  f(g(x)) = 4x^2 - 4
  \]
  So, \( f \circ g = 4x^2 - 4 \).

- **Find \( g \circ f \):**
  \[
  g(f(x)) = g(4x)
  \]
  Substitute \( f(x) \) into \( g(x) \):
  \[
  g(4x) = (4x)^2 - 1
  \]
  Simplify:
  \[
  g(f(x)) = 16x^2 - 1
  \]
  So, \( g \circ f = 16x^2 - 1 \).

### 3. \( f(x) = -x^2 + 1 \) and \( g(x) = x \)

- **Find \( f \circ g \):**
  \[
  f(g(x)) = f(x)
  \]
  Since \( g(x) = x \), this is just \( f(x) \):
  \[
  f(g(x)) = -x^2 + 1
  \]
  So, \( f \circ g = -x^2 + 1 \).

- **Find \( g \circ f \):**
  \[
  g(f(x)) = g(-x^2 + 1)
  \]
  Since \( g(x) = x \), this is simply:
  \[
  g(f(x)) = -x^2 + 1
  \]
  So, \( g \circ f = -x^2 + 1 \).

### Summary:
1. \( f(x) = 3x - 2 \), \( g(x) = \frac{1}{3}(x + 2) \)
   - \( f \circ g = x \)
   - \( g \circ f = x \)

2. \( f(x) = 4x \), \( g(x) = x^2 - 1 \)
   - \( f \circ g = 4x^2 - 4 \)
   - \( g \circ f = 16x^2 - 1 \)

3. \( f(x) = -x^2 + 1 \), \( g(x) = x \)
   - \( f \circ g = -x^2 + 1 \)
   - \( g \circ f = -x^2 + 1 \)

Would you like further details or have any questions? Here are 5 related questions to consider:

1. What is the general definition of function composition?
2. How do we verify if two functions are inverses using composition?
3. Can \( f \circ g \) and \( g \circ f \) be equal for all functions?
4. What happens to the domain and range when composing two functions?
5. Can you find the inverse functions of \( f(x) = 3x - 2 \) and \( g(x) = \frac{1}{3}(x + 2) \)?

**Tip:** When composing functions, always substitute one function entirely into the other, replacing every occurrence of \( x \).

find f . g and g . f. 1. f(x) = 3x - 2; g(x) = 1/3(x + 2) 2. f(x) = 4x; g(x) = x^2 - 1 3. f(x) = -x^2 + 1; g(x) = x

Learn how to solve function compositions for three different function pairs, including linear and quadratic functions. This problem involves composing functions f(x) and g(x) with detailed step-by-step explanations. Suitable for high school students (Grades 9-12).

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