To find the domain and range of the inverse function \( f^{-1}(x) \) given the function \( f(x) = \frac{1}{3x + 2} \), we can follow these steps:
Step 1: Find the Domain of \( f(x) \)
The domain of \( f(x) \) is determined by the values of \( x \) that keep the denominator from being zero.
Set the denominator equal to zero to find restrictions:
\[ 3x + 2 = 0 \]
Solving for \( x \):
\[ 3x = -2 \implies x = -\frac{2}{3} \]
Thus, the function \( f(x) \) is undefined at \( x = -\frac{2}{3} \). Therefore, the domain of \( f(x) \) is:
\[ \text{Domain of } f: \left( -\infty, -\frac{2}{3} \right) \cup \left( -\frac{2}{3}, \infty \right) \]
Step 2: Find the Range of \( f(x) \)
Next, let's determine the range of \( f(x) \).
The function \( f(x) = \frac{1}{3x + 2} \) approaches 0 as \( x \to -\infty \) or \( x \to \infty \), but it never reaches 0. The output can be any value except 0. Therefore, the range of \( f(x) \) is:
\[ \text{Range of } f: \left( -\infty, 0 \right) \cup \left( 0, \infty \right) \]
Step 3: Find the Domain and Range of the Inverse Function
Using the properties of inverse functions, we have:
- The domain of \( f^{-1}(x) \) is equal to the range of \( f(x) \).
- The range of \( f^{-1}(x) \) is equal to the domain of \( f(x) \).
Thus, we find:
- The domain of \( f^{-1}(x) \) is \( \left( -\infty, 0 \right) \cup \left( 0, \infty \right) \)
- The range of \( f^{-1}(x) \) is \( \left( -\infty, -\frac{2}{3} \right) \cup \left( -\frac{2}{3}, \infty \right) \)
Conclusion
Putting it all together, the correct response would be:
Domain of \( f^{-1}(x) \): \( \left( -\infty, 0 \right) \cup \left( 0, \infty \right) \)
Range of \( f^{-1}(x) \): \( \left( -\infty, -\frac{2}{3} \right) \cup \left( -\frac{2}{3}, \infty \right) \)
None of the provided options exactly match, as they list different domain and range sets. Please verify the options or details provided, or clarify if there was a misunderstanding.
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