What is the inverse of f(x)=(x^2)/(1+x^2) ?

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

inverse

f (x) = \frac{x ^{2}}{1 + x ^{2}}

- \frac{x}{x - 1}, - - \frac{x}{x - 1}

Solution steps

y = \frac{x ^{2}}{1 + x ^{2}}

Replace

x

with

y

x = \frac{y ^{2}}{1 + y ^{2}}

Solve for

y, x = \frac{y ^{2}}{1 + y ^{2}}

- \frac{x}{x - 1}, - - \frac{x}{x - 1}

f (x) = \frac{3 - 2 x}{x - 5}

f (x) = \frac{5 - 3 x}{2 x - 3}

f (x) = \frac{5 - 2 x}{4 x - 2}

f (x) = 2 (x - 3)^{2} + 5

f (x) = \frac{5 - x}{3 x - 4}

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