The one-to-one function \( h \) is defined below. \[ h(x)=\frac{x-7}{5 x+8} \] Find \( h^{-1}(x) \), where \( h^{-1} \) is the inverse of \( h \). Also state the domain and range of \( h^{-1} \) in interval notation. Domain of \( h^{-1}(x)=\square \)

To find the inverse function \( h^{-1}(x) \), we start by setting \( y = h(x) \): \[ y = \frac{x-7}{5x+8} \] Next, we swap \( x \) and \( y \): \[ x = \frac{y-7}{5y+8} \] Now, we solve for \( y \). Multiply both sides by \( 5y + 8 \): \[ x(5y + 8) = y - 7 \] Distributing gives us: \[ 5xy + 8x = y - 7 \] Rearranging terms to isolate \( y \): \[ 5xy - y = -8x - 7 \] Factoring out \( y \): \[ y(5x - 1) = -8x - 7 \] Now, divide both sides by \( (5x - 1) \): \[ y = \frac{-8x - 7}{5x - 1} \] Thus, we find the inverse function: \[ h^{-1}(x) = \frac{-8x - 7}{5x - 1} \] **Domain of \( h^{-1}(x) \)**: The function is undefined when the denominator is zero, so we need to solve: \[ 5x - 1 = 0 \implies x = \frac{1}{5} \] Thus, the domain of \( h^{-1}(x) \) is: \[ \text{Domain of } h^{-1}(x) = (-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty) \] **Range of \( h^{-1}(x) \)**: The range of the inverse function corresponds to the domain of the original function \( h(x) \). For \( h \), since the denominator can never be zero, we have: \[ 5x + 8 \neq 0 \implies x \neq -\frac{8}{5} \] The range is therefore: \[ \text{Range of } h^{-1}(x) = (-\infty, -\frac{8}{5}) \cup (-\frac{8}{5}, \infty) \] So to summarize: \[ h^{-1}(x) = \frac{-8x - 7}{5x - 1} \] \(\text{Domain of } h^{-1}(x) = (-\infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)\) and \(\text{Range of } h^{-1}(x) = (-\infty, -\frac{8}{5}) \cup (-\frac{8}{5}, \infty)\).