If f(x) = (x + 2) / (x + 3), which function is the inverse of f? 

a. 
(
1
+
x
)
(
1
−
x
)

b. 
(
3
x
−
2
)
(
1
−
x
)

c. 
(
3
x
−
1
)
(
x
−
2
)

d. 
(
3
x
−
2
)
(
1
+
x
)


To find the inverse of the function \( f(x) = \frac{x + 2}{x + 3} \), we follow these steps:

### Step 1: Replace \( f(x) \) with \( y \)
\[
y = \frac{x + 2}{x + 3}
\]

### Step 2: Solve for \( x \)
We need to express \( x \) in terms of \( y \). Start by multiplying both sides by \( x + 3 \) to eliminate the denominator:
\[
y(x + 3) = x + 2
\]
Expand:
\[
yx + 3y = x + 2
\]
Now, isolate \( x \). First, bring all terms involving \( x \) to one side and constants to the other:
\[
yx - x = 2 - 3y
\]
Factor out \( x \) on the left-hand side:
\[
x(y - 1) = 2 - 3y
\]
Finally, solve for \( x \):
\[
x = \frac{2 - 3y}{y - 1}
\]

### Step 3: Swap \( x \) and \( y \)
To find the inverse function, replace \( y \) with \( x \):
\[
f^{-1}(x) = \frac{2 - 3x}{x - 1}
\]

### Step 4: Compare with the given options
We see that this matches option **b**:
\[
f^{-1}(x) = \frac{3x - 2}{1 - x}
\]

Thus, the inverse function is **option b**.

Would you like more details on any of these steps or have further questions?

Here are 5 related questions to explore:
1. How can you verify if two functions are inverses of each other?
2. What is the domain of the function \( f(x) = \frac{x + 2}{x + 3} \)?
3. How do you find the domain and range of inverse functions?
4. What does it mean graphically for two functions to be inverses?
5. Can every function have an inverse? Why or why not?

**Tip:** To verify if two functions are inverses, compose them. If \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \), they are indeed inverses!

If f(x) = (x + 2) / (x + 3), which function is the inverse of f?

a. (1 + x)(1 − x)
b. (3x − 2)(1 − x)
c. (3x − 1)(x − 2)
d. (3x − 2)(1 + x)

Discover how to find the inverse of the function f(x) = (x + 2) / (x + 3) with a clear step-by-step solution. This problem explores algebraic concepts, including inverse functions and rational functions. Ideal for students in Grades 9-12.

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