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

The inverse of z/((z+1)^2) is (-2z+1+sqrt(-4z+1))/(2z),(-2z+1-sqrt(-4z+1))/(2z)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

\frac{- 2 z + 1 + - 4 z + 1}{2 z}, \frac{- 2 z + 1 - - 4 z + 1}{2 z}

Solution steps

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

Replace

z

with

y

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

Solve for

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

\frac{- 2 z + 1 + - 4 z + 1}{2 z}, \frac{- 2 z + 1 - - 4 z + 1}{2 z}

\frac{4 x ^{2} - 9}{2 x - 3}

y = 6 cos (2 π x)

- x^{2} + 20 x - 11

f (x) = 3 sin (2 x)

(0, 5), (- 3, 6)

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