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

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

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

inverse

\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}

(0, - 1), (- 5, - 2)

f (x) = 3 sin (4 x)

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

\frac{7}{3 x ^{2}}

f (x) = \frac{x + 1}{9}

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