What is the inverse of f(x)=x^2-4x+5,x>= 2 ?

The inverse of f(x)=x^2-4x+5,x>= 2 is 2+sqrt(x-1),2-sqrt(x-1)

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inverse of f(x)=x^2-4x+5,x>= 2

inverse

f (x) = x^{2} - 4 x + 5, x \geq 2

2 + x - 1, 2 - x - 1

Solution steps

y = x^{2} - 4 x + 5

Replace

x

with

y

x = y^{2} - 4 y + 5

Solve for

y, x = y^{2} - 4 y + 5

2 + x - 1, 2 - x - 1

7 x^{2} - 8

f (x) = (2 x - 4)^{2}

\frac{1 e ^{-} ( 2 s )}{( s ^{2} - s )}

7 x^{2} - 2

f (x) = - 5 x - 5, x \geq 5

What is the inverse of f(x)=x^2-4x+5,x>= 2 ?
The inverse of f(x)=x^2-4x+5,x>= 2 is 2+sqrt(x-1),2-sqrt(x-1)