What is the inverse of f(x)=1-sqrt(2x-x^2) ?

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

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inverse of f(x)=1-sqrt(2x-x^2)

inverse

f (x) = 1 - 2 x - x^{2}

1 - x (- x + 2), x (- x + 2) + 1

Solution steps

y = 1 - 2 x - x^{2}

Replace

x

with

y

x = 1 - 2 y - y^{2}

Solve for

y, x = 1 - 2 y - y^{2}

1 - x (- x + 2), x (- x + 2) + 1

f (x) = x^{2} + 3 x + 4

x^{2} + x - 2

f (x) = 2 x^{2} - 12 x + 13, x \leq 1

y = 37.75 x + 3.64

f (x) = x^{3 + 4}

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