If $f(x) = \frac{4x + 3}{6x - 4}$, $x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$ where $g : R - \{\frac{2}{3}\} \to R - \{\frac{2}{3}\}$, then $(g \circ g \circ g \circ g \circ g)(3) = $

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$. Then,the number of points in $R$ where $(f \circ g)(x)$ is $NOT$ differentiable is equal to

If function $f: R \rightarrow R$ is defined by $f(x) = (3 - x^5)^{\frac{1}{5}}$,then $(f \circ f)(x) = $ . . . . . . .

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

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:

If $f:[0,3] \rightarrow [0,3]$ is defined by $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$,then $f(f(x))$ is: