Sec 6.6 Inverse Trig: Probien 3 (1 point) Find the domain and the range of \( g(x)=\sin ^{-1}(3 x+1) \), Domain: \( \square \leq x \leq \square \) Range: \( \square \leq y \leq \square \) Note: You can earn partial credit on this problem.

To determine the domain of \( g(x) = \sin^{-1}(3x+1) \), we need to ensure that the argument of the inverse sine function, \( 3x + 1 \), falls within the valid output of the sine function, which is between -1 and 1. Therefore, we set up the inequalities: \[ -1 \leq 3x + 1 \leq 1. \] Solving these, we get: 1. For \( 3x + 1 \geq -1 \): \[ 3x \geq -2 \implies x \geq -\frac{2}{3}. \] 2. For \( 3x + 1 \leq 1 \): \[ 3x \leq 0 \implies x \leq 0. \] So the domain is: \[ -\frac{2}{3} \leq x \leq 0. \] Next, for the range, since the output of the function \( g(x) = \sin^{-1}(y) \) where \( y \) takes values from the domain, we know the range of the inverse sine function is between \(-\frac{\pi}{2}\) and \( \frac{\pi}{2}\): \[ -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}. \] Putting it all together, the answers are: Domain: \( -\frac{2}{3} \leq x \leq 0 \) Range: \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)