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

The inverse of (x^2-x+1)/(x^2+1) is (1+sqrt(-4x^2+8x-3))/(2(1-x)),(1-sqrt(-4x^2+8x-3))/(2(1-x))

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

inverse

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

\frac{1 + - 4 x ^{2} + 8 x - 3}{2 ( 1 - x )}, \frac{1 - - 4 x ^{2} + 8 x - 3}{2 ( 1 - x )}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + - 4 x ^{2} + 8 x - 3}{2 ( 1 - x )}, \frac{1 - - 4 x ^{2} + 8 x - 3}{2 ( 1 - x )}

\frac{8 x}{3}

y = \frac{60 x}{1 x ^{2} + 100}

21 x^{2} + 5 x - 84

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

f (x) = (x + 1)^{0.5}

What is the inverse of (x^2-x+1)/(x^2+1) ?
The inverse of (x^2-x+1)/(x^2+1) is (1+sqrt(-4x^2+8x-3))/(2(1-x)),(1-sqrt(-4x^2+8x-3))/(2(1-x))