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

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

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

inverse

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

\frac{1 + 1 - 4 x ^{2}}{2 x}, \frac{1 - 1 - 4 x ^{2}}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + 1 - 4 x ^{2}}{2 x}, \frac{1 - 1 - 4 x ^{2}}{2 x}

f (x) = - \frac{3}{2} x + \frac{3}{2}

6000 - 500 x

y = 0.5 x^{2} + 2

- 8 y = - 7 x + 20

f (x) = 4 e^{4 x}

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