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}

f (x) = 2 x^{2} - 8 x + 3

x - y = - 5

\frac{1}{x}

f (x) = \frac{9 - t ^{2}}{3 - t}

f (x) = \frac{3 x}{x ^{2} - 9}

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)