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

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

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

inverse

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

\frac{x + x ^{2} + 4 x}{2}, \frac{x - x ^{2} + 4 x}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x + x ^{2} + 4 x}{2}, \frac{x - x ^{2} + 4 x}{2}

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

2 x - 3 y = 10

h (x) = \frac{x + 1}{x ( x - 3 )}

- 1 - x^{5}

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