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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

\frac{1 + - x ( - x - 1 ) + 1}{x}, \frac{1 - - x ( - x - 1 ) + 1}{x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + - x ( - x - 1 ) + 1}{x}, \frac{1 - - x ( - x - 1 ) + 1}{x}

f (x) = \frac{1}{2} x - 9

y = lo g_{2} (x - 10)

y = (x - 1)^{2} + 2

f (x) = 2 (x - 1)^{2}

7 - x

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