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

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

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+x+1)/x

inverse

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

\frac{- 1 + x + x ^{2} - 2 x - 3}{2}, \frac{- 1 + x - x ^{2} - 2 x - 3}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 1 + x + x ^{2} - 2 x - 3}{2}, \frac{- 1 + x - x ^{2} - 2 x - 3}{2}

f (x) = 5 x^{2} + 8

f (x) = \frac{x}{3} - 1

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

f (x) = 3 (x + 4)^{2} - 2

f (x) = 8 x - 3

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