What is the inverse of f(x)=(x^2+1)/(x^2+3) ?

The inverse of f(x)=(x^2+1)/(x^2+3) is sqrt((1-3x)/(x-1)),-sqrt((1-3x)/(x-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of f(x)=(x^2+1)/(x^2+3)

inverse

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

\frac{1 - 3 x}{x - 1}, - \frac{1 - 3 x}{x - 1}

Solution steps

y = \frac{x ^{2} + 1}{x ^{2} + 3}

Replace

x

with

y

x = \frac{y ^{2} + 1}{y ^{2} + 3}

Solve for

y, x = \frac{y ^{2} + 1}{y ^{2} + 3}

\frac{1 - 3 x}{x - 1}, - \frac{1 - 3 x}{x - 1}

What is the inverse of f(x)=(x^2+1)/(x^2+3) ?
The inverse of f(x)=(x^2+1)/(x^2+3) is sqrt((1-3x)/(x-1)),-sqrt((1-3x)/(x-1))