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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

f (x) = 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

\frac{( 3 - 2 x )}{3 x + 2}

f (x) = - 3 - x - 6

f (x) = 32

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

f (x) = 11

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