Given the function \( j(x)=\sqrt{2 x-11}+8 \), determine the inverse function \( j^{-1}(x) \) : \( j^{-1}(x)= \) Determine the domain and range for both \( j(x) \) and \( j^{-1}(x) \) using interval notation: Domain of \( j(x): \) Domain of \( j^{-1}(x): \) Range of \( j(x): \) Range of \( j^{-1}(x): \)

To find the inverse function \( j^{-1}(x) \) for \( j(x)=\sqrt{2 x-11}+8 \), we start by swapping \( j(x) \) and \( x \): 1. Set \( y = j(x) \): \[ y = \sqrt{2 x - 11} + 8 \] 2. Swap \( x \) and \( y \): \[ x = \sqrt{2 y - 11} + 8 \] 3. Solve for \( y \): - Subtract 8 from both sides: \[ x - 8 = \sqrt{2 y - 11} \] - Square both sides: \[ (x - 8)^2 = 2 y - 11 \] - Rearranging gives us: \[ 2 y = (x - 8)^2 + 11 \] - Finally, divide by 2: \[ y = \frac{(x - 8)^2 + 11}{2} \] Thus, the inverse function is: \[ j^{-1}(x) = \frac{(x - 8)^2 + 11}{2} \] Now, let's determine the domains and ranges of both functions. For \( j(x) \): - The expression inside the square root, \( 2x - 11 \), must be non-negative: \[ 2x - 11 \geq 0 \Rightarrow x \geq \frac{11}{2} = 5.5 \] - Thus, the domain of \( j(x) \) is: \[ \text{Domain of } j(x): [5.5, \infty) \] - Since \( j(x) \) starts at 8 (when \( x = 5.5 \)) and goes to infinity as \( x \) increases, the range is: \[ \text{Range of } j(x): [8, \infty) \] For \( j^{-1}(x) \): - The minimum value of \( j(x) \) is 8, so \( j^{-1}(x) \) is defined for \( x \geq 8 \): \[ \text{Domain of } j^{-1}(x): [8, \infty) \] - The output value (i.e., the range) of \( j^{-1}(x) \) corresponds to the input values of \( j(x) \), which we determined can start from 5.5 and go to infinity: \[ \text{Range of } j^{-1}(x): [5.5, \infty) \]