Inverse function help

f(x) = x^2 - 2x + 5, x is greater or equal to 1
Find the inverse of the function

More importantly please show me how to get to the answer as well.

let y = x^2 - 2x + 5 (x >= 1, so y >=4 )

for inverse, x = y^2 - 2y + 5
solve for y ==>
y^2 - 2y + 5 - x = 0

y^2 - 2y + 1 + 4 - x = 0
y^2 - 2y + 1 = x - 4
(y - 1)^2 = x - 4
y - 1 = +/- sqrt(x - 4)
y = 1 +/- sqrt(x - 4)
but since the domain of the original function is x>= 1, that means the range of the inverse is y >= 1
thus that means take the positive root

y = 1 + sqrt(x - 4) <== inverse
(and as expected, the domain is x >= 4, matching the range of the original function

1) Replace f(x) with y
2) Swap y and x
3) Solve for y

f(x) = x²- 2x + 5
y = x² - 2x + 5
x = y² - 2y + 5
x - 5 = y² - 2y <---complete the square
x - 5 + 1 = y² - 2y + 1
x - 4 = (y - 1)²
±√(x - 4) = √(y - 1)²
±√(x - 4) = y - 1
1 ±√(x - 4) = y

Since the domain of f(x) is x ≥ 1, we want the positive version of f(x)^(-1) because,

f(x) = x² - 2x + 5, (2, 5)

5 = (2)² - 2(2) + 5

f(x)^(-1) = 1 + √(x - 4), (5, 2)

1 + √(5 - 4) = 2

answer---> f(x)^(-1) = 1 + √(x - 4)