Question

The function f left parenthesis x right parenthesis equals StartFraction x plus 4 Over x minus 2 EndFraction
is​ one-to-one. For the​ function,
a. Find an equation for f Superscript negative 1 Baseline left parenthesis x right parenthesis​, the inverse function.
b. Verify that your equation is correct by showing that f left parenthesis f Superscript negative 1 Baseline left parenthesis x right parenthesis right parenthesisequalsx and f Superscript negative 1 Baseline left parenthesis f left parenthesis x right parenthesis right parenthesisequalsx.
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Part 1
a. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
​(Simplify your answer. Use integers or fractions for any numbers in the​ expression.)
A.
f Superscript negative 1 Baseline left parenthesis x right parenthesisequals
  
enter your response here​, for xless than or equals
  
enter your response here
B.
f Superscript negative 1 Baseline left parenthesis x right parenthesisequals
  
enter your response here​, for xnot equals
  
enter your response here
C.
f Superscript negative 1 Baseline left parenthesis x right parenthesisequals
  
enter your response here​, for xgreater than or equals
  
enter your response here
D.
f Superscript negative 1 Baseline left parenthesis x right parenthesisequals
  
enter your response here​, for all x

Answer 1

To find the inverse of the function \( f(x) = \frac{x + 4}{x - 2} \), we will follow these steps:

### Part 1: Finding the Inverse Function

1. **Replace \( f(x) \) with \( y \)**:
\[
y = \frac{x + 4}{x - 2}
\]

2. **Switch \( x \) and \( y \)** (this is the key step when finding the inverse):
\[
x = \frac{y + 4}{y - 2}
\]

3. **Solve for \( y \)**:
- Multiply both sides by \( y - 2 \):
\[
x(y - 2) = y + 4
\]
- Distribute \( x \):
\[
xy - 2x = y + 4
\]
- Rearrange the equation to group terms with \( y \):
\[
xy - y = 2x + 4
\]
- Factor out \( y \) from the left side:
\[
y(x - 1) = 2x + 4
\]
- Divide by \( x - 1 \) (as long as \( x \neq 1 \)):
\[
y = \frac{2x + 4}{x - 1}
\]

4. **Write the inverse function**:
\[
f^{-1}(x) = \frac{2x + 4}{x - 1}
\]

### Answer for Part 1:
So, we have:
\[
f^{-1}(x) = \frac{2x + 4}{x - 1} \quad \text{for} \; x \neq 1
\]

### Part 2: Verifying the Inverse Function

We will verify \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

1. **Verify \( f(f^{-1}(x)) \)**:
\[
f(f^{-1}(x)) = f\left( \frac{2x + 4}{x - 1} \right) = \frac{\left( \frac{2x + 4}{x - 1} + 4\right)}{\left( \frac{2x + 4}{x - 1} - 2\right)}
\]
- Simplify the numerator:
\[
\frac{2x + 4}{x - 1} + 4 = \frac{2x + 4 + 4(x - 1)}{x - 1} = \frac{2x + 4 + 4x - 4}{x - 1} = \frac{6x}{x - 1}
\]

- Simplify the denominator:
\[
\frac{2x + 4}{x - 1} - 2 = \frac{2x + 4 - 2(x - 1)}{x - 1} = \frac{2x + 4 - 2x + 2}{x - 1} = \frac{6}{x - 1}
\]

- Therefore:
\[
f(f^{-1}(x)) = \frac{\frac{6x}{x - 1}}{\frac{6}{x - 1}} = x
\]

2. **Verify \( f^{-1}(f(x)) \)**:
\[
f^{-1}(f(x)) = f^{-1}\left( \frac{x + 4}{x - 2} \right) = \frac{2\left( \frac{x + 4}{x - 2} \right) + 4}{\left( \frac{x + 4}{x - 2} \right) - 1}
\]
- Simplify the numerator:
\[
2 \left( \frac{x + 4}{x - 2} \right) + 4 = \frac{2(x + 4) + 4(x - 2)}{x - 2} = \frac{2x + 8 + 4x - 8}{x - 2} = \frac{6x}{x - 2}
\]

- Simplify the denominator:
\[
\frac{x + 4}{x - 2} - 1 = \frac{x + 4 - (x - 2)}{x - 2} = \frac{6}{x - 2}
\]

- Therefore:
\[
f^{-1}(f(x)) = \frac{\frac{6x}{x - 2}}{\frac{6}{x - 2}} = x
\]

### Conclusion
Both verifications show that:
\[
f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x
\]

Hence, the calculations confirm that \( f^{-1}(x) = \frac{2x + 4}{x - 1} \) is indeed the inverse function.

**Final Answer** for Part 1:
\[
\text{B. } f^{-1}(x) = \frac{2x + 4}{x - 1}, \text{ for } x \neq 1
\]