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

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

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

inverse

y = \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{5 x + 1}{2 x + 3}

f (x) = \frac{x ^{2} + 1}{x ^{4} + 3 x ^{2} + 1}

y = 4 x^{2} + 34 x

f (x) = x (500 - 2 x)

f (x) = \frac{( x - 2 )}{( x + 7 )}

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