Unit 6 Radical Functions Homework 8 Inverse Relations And //free\\ ✦ Trusted Source

The original function ( f(x) = \sqrtx - 3 + 2 ) has a domain of ( x \ge 3 ) and a range of ( y \ge 2 ). When we square both sides in Step 4, we introduced extraneous solutions. Therefore, the inverse ( f^-1(x) = (x - 2)^2 + 3 ) is only valid for the restricted domain ( x \ge 2 ) (because the range of the original function becomes the domain of the inverse). Final Answer: ( f^-1(x) = (x - 2)^2 + 3, \quad x \ge 2 )