What is the inverse of f(x)=(1/(x+3))^2-2 ?

The inverse of f(x)=(1/(x+3))^2-2 is sqrt(1/(x+2))-3,-sqrt(1/(x+2))-3

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inverse of f(x)=(1/(x+3))^2-2

inverse

f (x) = (\frac{1}{x + 3})^{2} - 2

\frac{1}{x + 2} - 3, - \frac{1}{x + 2} - 3

Solution steps

y = (\frac{1}{x + 3})^{2} - 2

Replace

x

with

y

x = (\frac{1}{y + 3})^{2} - 2

Solve for

y, x = (\frac{1}{y + 3})^{2} - 2

\frac{1}{x + 2} - 3, - \frac{1}{x + 2} - 3

f (x) = (x + 7) mod 10

- 4 x^{2} - 8 x - 6

f (x) = - \frac{x}{2} + 3

f (x) = 5 x^{2} + 2, x \geq 0

y = 0.5 x + 2

What is the inverse of f(x)=(1/(x+3))^2-2 ?
The inverse of f(x)=(1/(x+3))^2-2 is sqrt(1/(x+2))-3,-sqrt(1/(x+2))-3