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

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

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

inverse

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

- \frac{x}{x - 1}, - - \frac{x}{x - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{x}{x - 1}, - - \frac{x}{x - 1}

f (x) = \frac{x}{( x + c )}

f (x) = \frac{x}{3} + 6

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

f (x) = \frac{100}{1 + 20 e ^{- 0.05 x}}

- \frac{3}{4} x + 6

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