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

The inverse of (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 (x^2+1)/(x^2+5)

inverse

\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}

f (x) = \frac{( ( - 2 x - 5 ) )}{( - 8 x + 5 )}

\frac{6 x - 7}{5 x + 9}

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

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

y = 2 (x - 2)^{3}

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