What is the inverse of f(x)=((x+1))/((x^2+4)) ?

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

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inverse of f(x)=((x+1))/((x^2+4))

inverse

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

\frac{1 + 1 - 4 x ( 4 x - 1 )}{2 x}, \frac{1 - 1 - 4 x ( 4 x - 1 )}{2 x}

Solution steps

y = \frac{x + 1}{x ^{2} + 4}

Replace

x

with

y

x = \frac{y + 1}{y ^{2} + 4}

Solve for

y, x = \frac{y + 1}{y ^{2} + 4}

\frac{1 + 1 - 4 x ( 4 x - 1 )}{2 x}, \frac{1 - 1 - 4 x ( 4 x - 1 )}{2 x}

9 - x^{2} - x + 2

f (x) = 2 x^{3} - 6

x - 4

f (x) = x - 2 y = - 4

- 6 + ln (x)

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