Answer
$f^{-1}(x)=\dfrac{1}{x}-2$
Work Step by Step
$f(x)=\dfrac{1}{x+2}$
Rewrite this expression as $y=\dfrac{1}{x+2}$ and solve for $x$:
$y=\dfrac{1}{x+2}$
Take $x+2$ to multiply the left side:
$(x+2)y=1$
Take $y$ to divide the right side:
$x+2=\dfrac{1}{y}$
Take $2$ to the right side:
$x=\dfrac{1}{y}-2$
Interchange $x$ and $y$:
$y=\dfrac{1}{x}-2$
The inverse of the initial function is $f^{-1}(x)=\dfrac{1}{x}-2$
