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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{7 x + 1}{x - 1}

- \frac{1}{2} x + 4

f (x) = 3 - 3 x, x \geq 0

f (x) = lo g_{2} (3 x + 4) - 5

f (x) = \frac{4 - x}{x}

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