Question

Find the domain and range of f superscript negative 1 baseline where lower f left parenthesis x right parenthesis equals Start Fraction 1 over 3 x plus 2 End Fraction.
(1 point)
Responses

domain: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
range: left parenthesis negative infinity comma negative Start Fraction 2 over 3 End Fraction right parenthesis union left parenthesis negative Start Fraction 2 over 3 End Fraction comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma negative Start Fraction 2 over 3 End Fraction right parenthesis union left parenthesis negative Start Fraction 2 over 3 End Fraction comma infinity right parenthesis

domain: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
range: left parenthesis negative infinity comma Start Fraction 2 over 3 End Fraction right parenthesis union left parenthesis Start Fraction 2 over 3 End Fraction comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma Start Fraction 2 over 3 End Fraction right parenthesis union left parenthesis Start Fraction 2 over 3 End Fraction comma infinity right parenthesis

domain: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
range: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: left parenthesis negative infinity comma negative Start Fraction 2 over 3 End Fraction right parenthesis union left parenthesis negative Start Fraction 2 over 3 End Fraction comma infinity right parenthesis
range: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma negative Start Fraction 2 over 3 End Fraction right parenthesis union left parenthesis negative Start Fraction 2 over 3 End Fraction comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

Answer 1

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.

1. **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:
\[
(-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty)
\]

2. **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:
\[
(-\infty, 0) \cup (0, \infty)
\]

3. **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:** \( (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty) \)