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

The inverse of f(z)=(z^2-z)/(z+1) is (1+z+sqrt(z^2+6z+1))/2 ,(1+z-sqrt(z^2+6z+1))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

\frac{1 + z + z ^{2} + 6 z + 1}{2}, \frac{1 + z - z ^{2} + 6 z + 1}{2}

Solution steps

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

Replace

z

with

y

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

Solve for

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

\frac{1 + z + z ^{2} + 6 z + 1}{2}, \frac{1 + z - z ^{2} + 6 z + 1}{2}

f (x) = 1 - \frac{1}{0.15 \cdot 223696201}

f (x) = 3 (3) - 5

f (x) = lo g_{2} (x + 1) - 3

3 x^{2} - x + 5

f (x) = lo g_{2} (x + 1) - 2

What is the inverse of f(z)=(z^2-z)/(z+1) ?
The inverse of f(z)=(z^2-z)/(z+1) is (1+z+sqrt(z^2+6z+1))/2 ,(1+z-sqrt(z^2+6z+1))/2