The one-to-one function \( f \) is defined below. \( f(x)=\frac{-5 x+2}{5+6 x} \) Find \( f^{-1}(x) \), where \( f^{-1} \) is the inverse of \( f \). Also state the domain and range of \( f^{-1} \) in interval notation. Domain of \( f^{-1}(x)= \) Range of \( f \)

**Step 1. Write the function equation with \( y \) in place of \( f(x) \):** \[ y = \frac{-5x+2}{5+6x} \] **Step 2. Solve for \( x \) in terms of \( y \).** Multiply both sides by \( 5+6x \) to eliminate the denominator: \[ y(5+6x) = -5x+2 \] Expand the left-hand side: \[ 5y + 6xy = -5x + 2 \] **Step 3. Collect the \( x \) terms on one side.** Bring the term \( -5x \) to the left-hand side: \[ 6xy + 5x = 2 - 5y \] Factor out \( x \) from the left-hand side: \[ x(6y+5) = 2-5y \] Assuming \( 6y+5 \neq 0 \), solve for \( x \): \[ x = \frac{2-5y}{6y+5} \] **Step 4. Write the inverse function.** Since we solved for \( x \) in terms of \( y \), the inverse function \( f^{-1}(y) \) is given by: \[ f^{-1}(y)=\frac{2-5y}{6y+5} \] Replacing \( y \) with \( x \) for the inverse function notation, we obtain: \[ f^{-1}(x)=\frac{2-5x}{6x+5} \] **Step 5. Determine the domain of \( f^{-1} \) and the range of \( f \).** The expression for \( f^{-1}(x) \) has a denominator \( 6x+5 \). We must have: \[ 6x+5 \neq 0 \quad \Longrightarrow \quad x\neq -\frac{5}{6} \] Thus, the domain of \( f^{-1} \) is: \[ (-\infty, -\frac{5}{6}) \cup (-\frac{5}{6}, \infty) \] Since the inverse function \( f^{-1} \) has as its domain the range of \( f \), the range of \( f \) is also: \[ (-\infty, -\frac{5}{6}) \cup (-\frac{5}{6}, \infty) \] **Final Answer:** \[ f^{-1}(x)=\frac{2-5x}{6x+5} \] Domain of \( f^{-1}(x)=(-\infty, -\frac{5}{6}) \cup (-\frac{5}{6}, \infty) \) Range of \( f \)=\((-\infty, -\frac{5}{6}) \cup (-\frac{5}{6}, \infty) \)