To find the domain and range of the inverse function \( f^{-1} \) for \( f(x) = \frac{1}{3x + 2} \), we will first determine the function's domain and range, and then use those to find the inverse's domain and range.
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Finding the domain of \( f(x) \):
The function \( f(x) = \frac{1}{3x + 2} \) is undefined when the denominator equals zero: \[ 3x + 2 = 0 \implies 3x = -2 \implies x = -\frac{2}{3} \] Thus, the domain of \( f(x) \) is: \[ (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty) \]
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Finding the range of \( f(x) \):
The function approaches zero as \( x \) approaches \(-\frac{2}{3}\) from left or right, but never actually reaches zero. The function \( f(x) \) decreases without bound as \( x \) approaches \(-\frac{2}{3}\). Hence, the range of \( f(x) \) includes all real numbers except for zero: \[ (-\infty, 0) \cup (0, \infty) \]
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Finding the domain and range of the inverse \( f^{-1}(x) \):
The domain of the inverse function \( f^{-1}(x) \) is equal to the range of \( f(x) \), while the range of \( f^{-1}(x) \) is equal to the domain of \( f(x) \).
Therefore, we have:
- Domain of \( f^{-1}(x) = (-\infty, 0) \cup (0, \infty) \)
- Range of \( f^{-1}(x) = (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty) \)
Thus, the correct response is:
- Domain: \( (-\infty, 0) \cup (0, \infty) \)
- Range: \( (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty) \)
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