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

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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) = \frac{3 x - 1}{4 x + 5}

f (x) = 2 x - (\frac{4}{3})

f (x) = (x - 14)^{2}, [14, \infty]

\frac{s ^{2}}{s ^{2} + 1}

f (x) = \frac{- 5 x + 4}{3 x - 1}

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