Determine the domain and the range of the function. \[ f(x)=\sin ^{-1} x+\cos ^{-1} x \]

Step 1 ：The function given is \(f(x)=\sin ^{-1} x+\cos ^{-1} x\).

Step 2 ：The domain of a function is the set of all possible input values, which produce a valid output from a particular function. The range of a function is the possible values of the output, or the dependent variable.

Step 3 ：For the function \(f(x)=\sin ^{-1} x+\cos ^{-1} x\), we need to consider the domains of the inverse sine and cosine functions separately.

Step 4 ：The domain of \(\sin ^{-1} x\) is \([-1, 1]\) and the domain of \(\cos ^{-1} x\) is also \([-1, 1]\). Therefore, the domain of the function \(f(x)=\sin ^{-1} x+\cos ^{-1} x\) is \([-1, 1]\).

Step 5 ：The range of \(\sin ^{-1} x\) is \([-π/2, π/2]\) and the range of \(\cos ^{-1} x\) is \([0, π]\). However, we are adding these two functions together, so we need to consider the minimum and maximum values that could result from this.

Step 6 ：The minimum value of \(\sin ^{-1} x + \cos ^{-1} x\) occurs when \(\sin ^{-1} x\) is at its minimum and \(\cos ^{-1} x\) is at its minimum. This would be \(-π/2 + 0 = -π/2\).

Step 7 ：The maximum value of \(\sin ^{-1} x + \cos ^{-1} x\) occurs when \(\sin ^{-1} x\) is at its maximum and \(\cos ^{-1} x\) is at its maximum. This would be \(π/2 + π = 3π/2\).

Step 8 ：Therefore, the range of the function \(f(x)=\sin ^{-1} x+\cos ^{-1} x\) is \([-π/2, 3π/2]\).

Step 9 ：\(\boxed{\text{The domain of the function } f(x)=\sin ^{-1} x+\cos ^{-1} x \text{ is } [-1, 1] \text{ and the range is } [-π/2, 3π/2]}\)