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

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

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inverse of y=sqrt(2x-x^2)

inverse

y = 2 x - x^{2}

1 - - x^{2} + 1, - x^{2} + 1 + 1

Solution steps

y = 2 x - x^{2}

Replace

x

with

y

x = 2 y - y^{2}

Solve for

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

1 - - x^{2} + 1, - x^{2} + 1 + 1

f (x) = \frac{( e ^{x} - 1 )}{e ^{x} + 1}

coth (x)

5 + 2 e^{- t}

f (x) = 3 \cdot x

f (2) = 3^{x + 2} + 1

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