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 \leq 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

f (x) = 7 x + 3

f (x) = x + 7 + 5

f (x) = \frac{5}{x + 8}

f (x) = - 2 sin (x - π)

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)