To find the inverse of the function $f(x) = \frac{x + 2}{x - 5}$, let’s go through the steps to find the inverse, determine the domain and range of both $f$ and $f^{-1}$, and then discuss how they would be graphed.

Step 1: Find the Inverse Function

1. Rewrite $f(x)$ as $y$: $y = \frac{x + 2}{x - 5}$

2. Swap $x$ and $y$ to find $f^{-1}(x)$: $x = \frac{y + 2}{y - 5}$

3. Solve for $y$: Multiply both sides by $y - 5$: $x(y - 5) = y + 2$ Distribute $x$: $xy - 5x = y + 2$ Move terms involving $y$ to one side: $xy - y = 5x + 2$ Factor out $y$: $y(x - 1) = 5x + 2$ Divide by $x - 1$ to solve for $y$: $y = \frac{5x + 2}{x - 1}$

So, the inverse function is: $f^{-1}(x) = \frac{5x + 2}{x - 1}$

Step 2: Determine the Domain and Range

For the function $f(x) = \frac{x + 2}{x - 5}$:

• Domain of $f$: The function $f(x)$ is undefined when the denominator $x - 5 = 0$, which gives $x = 5$. Thus, the domain of $f$ is all real numbers except $x = 5$: $\text{Domain of } f: (-\infty, 5) \cup (5, \infty)$

• Range of $f$: The function $f(x)$ approaches all real values as $x$ varies, but it cannot equal 1 (since $f(x) = 1$ would imply $x = 5$, which is excluded from the domain). Therefore, the range of $f$ is all real numbers except 1: $\text{Range of } f: (-\infty, 1) \cup (1, \infty)$

For the inverse function $f^{-1}(x) = \frac{5x + 2}{x - 1}$:

• Domain of $f^{-1}$: Similar reasoning applies, where the inverse function is undefined for $x = 1$. Thus, the domain of $f^{-1}$ is: $\text{Domain of } f^{-1}: (-\infty, 1) \cup (1, \infty)$

• Range of $f^{-1}$: The range of $f^{-1}$ will match the domain of $f$, so it is: $\text{Range of } f^{-1}: (-\infty, 5) \cup (5, \infty)$

Step 3: Graphing $f$ and $f^{-1}$

To graph $f(x)$ and $f^{-1}(x)$ on the same set of axes, you would plot both functions. They will be symmetric about the line $y = x$, as this line reflects a function and its inverse.

Would you like to see a plot of these functions, or do you need further clarification on any step?

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Here are 5 follow-up questions to deepen understanding:

1. Why is the line $y = x$ the axis of symmetry for a function and its inverse?
2. What would change if we shifted $f(x)$ horizontally or vertically?
3. How can we verify that $f^{-1}(x)$ we found is indeed the inverse of $f(x)$?
4. How would the domain and range change if $f(x)$ had an additional term, such as $f(x) = \frac{x + 2}{x - 5} + 3$?
5. What types of real-world problems involve finding the inverse of a function?

Tip: Remember that a function and its inverse will "undo" each other, meaning $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. This is a good way to check if you've found the correct inverse.