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

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

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

inverse

f (x) = \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}

lo g_{10} (1 - x^{2})

f (x) = 0.5 x^{2} - 2 x - 2

f (x) = ln (\frac{( x - 9 )}{x - 3})

x e^{x}

y (θ) = sin (θ + \frac{π}{2})

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