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The assignment asks for the inverse functions of the given functions. Let’s work through each one of them:

### 1. \( g(x) = \frac{x - 5}{x + 4} \)

To find the inverse, follow these steps:
1. Replace \( g(x) \) with \( y \):  
   \( y = \frac{x - 5}{x + 4} \)
2. Swap \( x \) and \( y \):  
   \( x = \frac{y - 5}{y + 4} \)
3. Solve for \( y \):  
   Multiply both sides by \( y + 4 \):  
   \( x(y + 4) = y - 5 \)  
   \( xy + 4x = y - 5 \)  
   Get all terms involving \( y \) on one side:  
   \( xy - y = -4x - 5 \)  
   Factor out \( y \):  
   \( y(x - 1) = -4x - 5 \)  
   Finally, solve for \( y \):  
   \( y = \frac{-4x - 5}{x - 1} \)

So the inverse is:
\[
g^{-1}(x) = \frac{-4x - 5}{x - 1}
\]

---

### 2. \( f(x) = 4(x - 5) + 3 \)

1. Replace \( f(x) \) with \( y \):  
   \( y = 4(x - 5) + 3 \)
2. Swap \( x \) and \( y \):  
   \( x = 4(y - 5) + 3 \)
3. Solve for \( y \):  
   \( x = 4y - 20 + 3 \)  
   \( x = 4y - 17 \)  
   Add 17 to both sides:  
   \( x + 17 = 4y \)  
   Divide by 4:  
   \( y = \frac{x + 17}{4} \)

So the inverse is:
\[
f^{-1}(x) = \frac{x + 17}{4}
\]

---

### 3. \( f(x) = 6 \sqrt{x - 9} \)

1. Replace \( f(x) \) with \( y \):  
   \( y = 6 \sqrt{x - 9} \)
2. Swap \( x \) and \( y \):  
   \( x = 6 \sqrt{y - 9} \)
3. Solve for \( y \):  
   Divide both sides by 6:  
   \( \frac{x}{6} = \sqrt{y - 9} \)  
   Square both sides:  
   \( \left( \frac{x}{6} \right)^2 = y - 9 \)  
   \( \frac{x^2}{36} = y - 9 \)  
   Add 9 to both sides:  
   \( y = \frac{x^2}{36} + 9 \)

So the inverse is:
\[
f^{-1}(x) = \frac{x^2}{36} + 9
\]

---

### 4. \( h(x) = \frac{2x - 9}{x} \)

1. Replace \( h(x) \) with \( y \):  
   \( y = \frac{2x - 9}{x} \)
2. Swap \( x \) and \( y \):  
   \( x = \frac{2y - 9}{y} \)
3. Solve for \( y \):  
   Multiply both sides by \( y \):  
   \( x y = 2y - 9 \)  
   Rearrange:  
   \( x y - 2y = -9 \)  
   Factor out \( y \):  
   \( y(x - 2) = -9 \)  
   Divide by \( (x - 2) \):  
   \( y = \frac{-9}{x - 2} \)

So the inverse is:
\[
h^{-1}(x) = \frac{-9}{x - 2}
\]

---

### 5. \( m(x) = \frac{x - 5}{x + 1} \)

1. Replace \( m(x) \) with \( y \):  
   \( y = \frac{x - 5}{x + 1} \)
2. Swap \( x \) and \( y \):  
   \( x = \frac{y - 5}{y + 1} \)
3. Solve for \( y \):  
   Multiply both sides by \( y + 1 \):  
   \( x(y + 1) = y - 5 \)  
   \( xy + x = y - 5 \)  
   Get all terms involving \( y \) on one side:  
   \( xy - y = -x - 5 \)  
   Factor out \( y \):  
   \( y(x - 1) = -x - 5 \)  
   Finally, solve for \( y \):  
   \( y = \frac{-x - 5}{x - 1} \)

So the inverse is:
\[
m^{-1}(x) = \frac{-x - 5}{x - 1}
\]

---

Would you like any further details or have any questions?

Here are 5 follow-up questions you might explore:
1. How can you verify that the inverse function you found is correct?
2. What are the domain and range restrictions of each inverse function?
3. Why do some inverse functions have restrictions on the domain that the original function did not?
4. How does the process of finding an inverse change for functions involving square roots?
5. Can every function have an inverse?

**Tip:** Always check if the function is one-to-one (i.e., passes the horizontal line test) before attempting to find the inverse.

Find the inverse functions of the given functions: 
1) g(x) = (x - 5)/(x + 4) 
2) f(x) = 4(x - 5) + 3 
3) f(x) = 6√(x - 9) 
4) h(x) = (2x - 9)/x 
5) m(x) = (x - 5)/(x + 1)

This problem involves finding inverse functions for different types of equations, including rational, linear, and square root functions. Learn the step-by-step process to solve for the inverse, using algebraic manipulation and key inverse function concepts. Suitable for students in Grades 11-12.

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