What is the inverse of (2z^2+1)/(z^2+1) ?

The inverse of (2z^2+1)/(z^2+1) is sqrt((1-z)/(z-2)),-sqrt((1-z)/(z-2))

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inverse of (2z^2+1)/(z^2+1)

inverse

\frac{2 z ^{2} + 1}{z ^{2} + 1}

\frac{1 - z}{z - 2}, - \frac{1 - z}{z - 2}

Solution steps

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

Replace

z

with

y

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

Solve for

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

\frac{1 - z}{z - 2}, - \frac{1 - z}{z - 2}

\frac{- 2 x - 9}{- 6 x - 7}

\frac{1}{1 + 2 I}

5 x + 15

f (x) = 4 - e^{- 5 x}

f (x) = x + \frac{12}{x}

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