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

The inverse of f(x)=(x^2)/(x+1) is (x+sqrt(x^2+4x))/2 ,(x-sqrt(x^2+4x))/2

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)/(x+1)

inverse

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

\frac{x + x ^{2} + 4 x}{2}, \frac{x - x ^{2} + 4 x}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x + x ^{2} + 4 x}{2}, \frac{x - x ^{2} + 4 x}{2}

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

g (x) = (x + 1)^{2}

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

f (x) = 4^{0.02 x}

y = \frac{5}{5 x}

What is the inverse of f(x)=(x^2)/(x+1) ?
The inverse of f(x)=(x^2)/(x+1) is (x+sqrt(x^2+4x))/2 ,(x-sqrt(x^2+4x))/2