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)

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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}

[2, - 3, - 5, 2]

f (x) = \frac{( x + 2 )}{( x + 7 )}

f (x) = 420 + 0.75 n

y = 1.6 x + 1.9 (50 - x)

\frac{d}{d} arctan (7 x + 10)

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)