What is the inverse of y= 1/((x+1)^2) ?

The inverse of y= 1/((x+1)^2) is sqrt(1/x)-1,-sqrt(1/x)-1

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inverse of y= 1/((x+1)^2)

inverse

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

\frac{1}{x} - 1, - \frac{1}{x} - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1}{x} - 1, - \frac{1}{x} - 1

5 x - 3

f (x) = 2 \cdot 11 - x

- 212

13.5 - 5 t

x = - 13.528 {3.38, 9 \leq y \leq 4.703}

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