What is the inverse of f(x)=(x^2+1)/(x^2+5) ?

The inverse of f(x)=(x^2+1)/(x^2+5) is sqrt((1-5x)/(x-1)),-sqrt((1-5x)/(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+5)

inverse

f (x) = \frac{x ^{2} + 1}{x ^{2} + 5}

\frac{1 - 5 x}{x - 1}, - \frac{1 - 5 x}{x - 1}

Solution steps

y = \frac{x ^{2} + 1}{x ^{2} + 5}

Replace

x

with

y

x = \frac{y ^{2} + 1}{y ^{2} + 5}

Solve for

y, x = \frac{y ^{2} + 1}{y ^{2} + 5}

\frac{1 - 5 x}{x - 1}, - \frac{1 - 5 x}{x - 1}

3 x^{2} + 4 x + 7

e^{- 12 x^{2}}

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

f (x) = x - 5, x \geq 5

\frac{D}{D x} 4 x + 10 x^{15}

What is the inverse of f(x)=(x^2+1)/(x^2+5) ?
The inverse of f(x)=(x^2+1)/(x^2+5) is sqrt((1-5x)/(x-1)),-sqrt((1-5x)/(x-1))