What is the inverse of f(z)=(4z^2)/((z-3)(3z+1)) ?

The inverse of f(z)=(4z^2)/((z-3)(3z+1)) is (4z+sqrt(z(25z-12)))/(3z-4),(4z-sqrt(z(25z-12)))/(3z-4)

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inverse of f(z)=(4z^2)/((z-3)(3z+1))

inverse

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

\frac{4 z + z ( 25 z - 12 )}{3 z - 4}, \frac{4 z - z ( 25 z - 12 )}{3 z - 4}

Solution steps

y = \frac{4 z ^{2}}{( z - 3 ) ( 3 z + 1 )}

Replace

z

with

y

z = \frac{4 y ^{2}}{( y - 3 ) ( 3 y + 1 )}

Solve for

y, z = \frac{4 y ^{2}}{( y - 3 ) ( 3 y + 1 )}

\frac{4 z + z ( 25 z - 12 )}{3 z - 4}, \frac{4 z - z ( 25 z - 12 )}{3 z - 4}

What is the inverse of f(z)=(4z^2)/((z-3)(3z+1)) ?
The inverse of f(z)=(4z^2)/((z-3)(3z+1)) is (4z+sqrt(z(25z-12)))/(3z-4),(4z-sqrt(z(25z-12)))/(3z-4)