The function f left parenthesis x right parenthesis equals 8 x minus 2 is​ one-to-one. 
​(a) Find the inverse of f and check the answer.
​(b) Find the domain and the range of f and f Superscript negative 1.
​(c) Graph​ f, f Superscript negative 1​, and yequalsx on the same coordinate axes.
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Part 1
​(a) f Superscript negative 1 Baseline left parenthesis x right parenthesisequals
  
enter your response here
​(Simplify your answer. Use integers or fractions for any numbers in the​ expression.)


### (a) Find the inverse of \( f(x) = 8x - 2 \).

To find the inverse, follow these steps:

1. **Replace** \( f(x) \) with \( y \):
   \[
   y = 8x - 2
   \]
2. **Swap** \( x \) and \( y \) to find the inverse function:
   \[
   x = 8y - 2
   \]
3. **Solve** for \( y \) (the inverse function):
   \[
   x + 2 = 8y
   \]
   \[
   y = \frac{x + 2}{8}
   \]
   So, the inverse function is:
   \[
   f^{-1}(x) = \frac{x + 2}{8}
   \]

4. **Check the answer**:
   To verify, we need to show that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
   - First, substitute \( f^{-1}(x) \) into \( f(x) \):
     \[
     f\left( f^{-1}(x) \right) = f\left( \frac{x + 2}{8} \right) = 8 \cdot \frac{x + 2}{8} - 2 = x + 2 - 2 = x
     \]
   - Now, substitute \( f(x) \) into \( f^{-1}(x) \):
     \[
     f^{-1}(f(x)) = f^{-1}(8x - 2) = \frac{(8x - 2) + 2}{8} = \frac{8x}{8} = x
     \]
   Since both hold true, the inverse is correct.

Thus, the inverse function is:
\[
f^{-1}(x) = \frac{x + 2}{8}
\]

### (b) Find the domain and range of \( f(x) \) and \( f^{-1}(x) \).

- The **domain of \( f(x) \)** is all real numbers \( \mathbb{R} \) because \( 8x - 2 \) is a linear function, which is defined for all real numbers.
  \[
  \text{Domain of } f(x) = \mathbb{R}
  \]
  
- The **range of \( f(x) \)** is also all real numbers \( \mathbb{R} \), since as \( x \) takes any real value, \( 8x - 2 \) can output any real number.
  \[
  \text{Range of } f(x) = \mathbb{R}
  \]

- The **domain of \( f^{-1}(x) \)** is the range of \( f(x) \), which is \( \mathbb{R} \).
  \[
  \text{Domain of } f^{-1}(x) = \mathbb{R}
  \]
  
- The **range of \( f^{-1}(x) \)** is the domain of \( f(x) \), which is \( \mathbb{R} \).
  \[
  \text{Range of } f^{-1}(x) = \mathbb{R}
  \]

### (c) Graph \( f(x) \), \( f^{-1}(x) \), and \( y = x \) on the same coordinate axes.

To sketch the graphs:

1. **Graph of \( f(x) = 8x - 2 \)**: This is a straight line with a slope of 8 and a y-intercept at -2.
   
2. **Graph of \( f^{-1}(x) = \frac{x + 2}{8} \)**: This is a straight line with a slope of \( \frac{1}{8} \) and a y-intercept at \( \frac{2}{8} = 0.25 \).

3. **Graph of \( y = x \)**: This is the line that passes through the origin with a slope of 1, serving as the reflection line for the function and its inverse.

Would you like me to generate a graph of these functions?

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### Relative Questions:
1. How do you verify if a function is one-to-one?
2. What conditions are necessary for a function to have an inverse?
3. How do you determine the domain and range of composite functions?
4. Can a quadratic function have an inverse? Why or why not?
5. How do you interpret the graph of \( y = x \) when comparing a function and its inverse?

### Tip:
When finding an inverse, always check your result by composing the function and its inverse to ensure that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

The function f(x) = 8x - 2 is one-to-one. (a) Find the inverse of f and check the answer. (b) Find the domain and the range of f and f⁻¹. (c) Graph f, f⁻¹, and y = x on the same coordinate axes.

Learn how to find the inverse of the function f(x) = 8x - 2, check your answer, and determine the domain and range of both f(x) and f⁻¹(x). Explore key concepts in algebra, including one-to-one functions, inverse functions, and graphing. Suitable for students in Grades 10-12.

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