91Ó°ÊÓ

We replace $$f(x)$$ with $$y$$ to rewrite the given function as: $$y = \frac{2}{x^2 + 2}$$

To find the inverse, we need to solve for $$x$$ in terms of $$y$$. Let's start by cross-multiplying: $$yx^2 + 2y = 2$$ Now, isolate the $$x^2$$ term: $$yx^2 = 2 - 2y$$ Divide both sides by y to make the equation with x^2 alone: $$x^2 = \frac{2 - 2y}{y}$$ Now, square root both sides to solve for x: $$x = \pm \sqrt{\frac{2 - 2y}{y}}$$

Now that we've solved for $$x$$, we can rewrite the inverse function as: $$f^{-1}(y) = \pm \sqrt{\frac{2 - 2y}{y}}$$

The domain of the inverse function consists of all the values of $$y$$ for which the function is defined. Since we have a square root, we need to consider that the expression inside the square root should always be non-negative (i.e. greater than or equal to 0), also there is a y value in the denominator that should not be zero. Let's consider the expression inside the square root: $$\frac{2 - 2y}{y} \geq 0$$ Since the denominator is positive for all non-zero y, we should only look at $$2 - 2y \geq 0$$. Now, let's solve for y: $$2 - 2y \geq 0 \Rightarrow y \leq 1$$ Since y should not be zero and it is in the denominator: $$y > 0$$ So, the domain of the inverse function is $$0 < y \leq 1$$. #Answer# The inverse function is $$f^{-1}(y) = \pm \sqrt{\frac{2 - 2y}{y}}$$ with a domain of $$0 < y \leq 1$$.