If $f: R \rightarrow R$ is defined by $f(x) = \frac{x}{x^{2}+1}$, find $f(f(2))$.

If $f: R \rightarrow R$ is defined by $f(x) = (3 - x^3)^{\frac{1}{3}}$,then $f \circ (f \circ f)(x) = $ . . . . . . .

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$. Then for all values of $x$,the value of $f(g(x))$ is:

$f(x) = (20 - x^4)^{1/4}$ for $0 < x < \sqrt{5}$, then $f(f(1/2))$ is equal to

If the function $f(x) = \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right)$ for $-1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$,then the domain of the composite function $(g \circ f)(x)$ is:

$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=2x-3$ and $g(x)=x^3+5$. Then,$(f \circ g)^{-1}(-9)$ is