Sure, let's go through the process to find the domain and range of the inverse function [tex]\( f^{-1} \)[/tex] for the given function [tex]\( f(x) = \frac{1}{3x + 2} \)[/tex].
### Step 1: Find [tex]\( f^{-1}(x) \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex], we start by setting [tex]\( f(x) \)[/tex] equal to [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1}{3x + 2} \][/tex]
Next, we solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y(3x + 2) = 1 \][/tex]
[tex]\[ 3xy + 2y = 1 \][/tex]
[tex]\[ 3xy = 1 - 2y \][/tex]
[tex]\[ x = \frac{1 - 2y}{3y} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(y) = \frac{1 - 2y}{3y} \][/tex]
### Step 2: Determine the Domain of [tex]\( f^{-1}(x) \)[/tex]
The domain of [tex]\( f \)[/tex] should reflect the range of [tex]\( f^{-1} \)[/tex]. For [tex]\( f(x) = \frac{1}{3x + 2} \)[/tex], the function is defined for all [tex]\( x \)[/tex] except where the denominator is zero. We solve for when the denominator is zero:
[tex]\[ 3x + 2 = 0 \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
Hence, [tex]\( f(x) \)[/tex] is defined for all [tex]\( x \in \mathbb{R} \)[/tex] except [tex]\( x = -\frac{2}{3} \)[/tex]. The range of [tex]\( f(x) \)[/tex] is therefore all real numbers except [tex]\( 0 \)[/tex] because as [tex]\( x \)[/tex] approaches [tex]\(-\frac{2}{3}\)[/tex], the function approaches infinity. Therefore, the range of [tex]\( f(x) \)[/tex] is [tex]\( \mathbb{R} \setminus \{0\} \)[/tex].
Since the range of [tex]\( f(x) \)[/tex] becomes the domain of [tex]\( f^{-1}(x) \)[/tex], the domain of [tex]\( f^{-1} \)[/tex] is [tex]\( \mathbb{R} \setminus \{0\} \)[/tex].
### Step 3: Determine the Range of [tex]\( f^{-1}(x) \)[/tex]
The range of [tex]\( f \)[/tex] becomes the domain of [tex]\( f^{-1} \)[/tex]. Therefore, the domain of [tex]\( f \)[/tex] (all real numbers except when [tex]\( 3x + 2 = 0 \)[/tex]) will be the range of [tex]\( f^{-1} \)[/tex]. That is:
[tex]\[ 3x + 2 \neq 0 \][/tex]
So, the domain of [tex]\( f \)[/tex], and hence the range of [tex]\( f^{-1} \)[/tex], is all real numbers ([tex]\( \mathbb{R} \)[/tex]).
### Conclusion
From the steps above:
- The inverse function is [tex]\( f^{-1}(x) = \frac{1 - 2x}{3x} \)[/tex].
- The domain of [tex]\( f^{-1}(x) \)[/tex] is [tex]\( \mathbb{R} \setminus \{0\} \)[/tex].
- The range of [tex]\( f^{-1}(x) \)[/tex] is all real numbers ([tex]\( \mathbb{R} \)[/tex]).
Therefore:
[tex]\[
f^{-1}(y) = \frac{1 - 2y}{3y}
\][/tex]
Domain of [tex]\( f^{-1} \)[/tex]: [tex]\(\mathbb{R}\)[/tex]
Range of [tex]\( f^{-1} \)[/tex]: [tex]\( \mathbb{R} \setminus \{0\} \)[/tex]
