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Answer:
(b) x ≥ 1; f⁻¹(x) = 1+√(x+1)
Step-by-step explanation:
We can find the inverse function by solving for y:
x = f(y)
x = y² -2y
x +1 = y² -2y +1
x +1 = (y -1)²
√(x +1) = y -1
1 +√(x +1) = y
So, the inverse function is ...

For this, the root will be non-negative, so the minimum value of this function is 1. That is, the minimum value of x for which the original function has this as an inverse is 1.
The domain restriction is x ≥ 1.
The inverse function is f⁻¹(x) = 1+√(x+1)