Answer
$f(x)=$\begin{cases}
\sqrt{1-x^2} & |x|\le2\\
|\frac{2x}{3}|-3 & |x|>2 \\
\end{cases}
Work Step by Step
Every image of x less than 2 or greater than 2 is equal to the graph of the function $f(x) = |\frac{2x}{3}|-3$, which is the absolute value of the line $y = \frac{2x}{3}$ displaced three units downwards (subtract 3), and every image of x in the interval [-2,2] is equal to the graph of the function $g(x) = \sqrt{1-x^2}$, which is the upper half of the circle of radius 2 centered at the origin (given by $x^2+y^2=4$). See the attached figure.
