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

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

inverse

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

(x + 2)^{5}

y = \frac{- 4 x ^{2} + 28 x + 31}{16}

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

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