If $g(x) = 1 - \sqrt{x}$ and $f(g(x)) = 5 + 4\sqrt{x} + x$, then the value of $f(6)$ is...

For $x \in R - \{0, 1\}$,let ${f_1}(x) = \frac{1}{x}$,${f_2}(x) = 1 - x$,and ${f_3}(x) = \frac{1}{1 - x}$ be three given functions. If a function $J(x)$ satisfies $(f_2 \circ J \circ f_1)(x) = f_3(x)$,then $J(x)$ is equal to:

If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

If $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$,then the range of $(f \circ g)(x)$ is:

If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$, then $(f \circ g)^{\prime}(1)=$

If $f(x) = \frac{2x - 3}{3x - 2}$ and $f_n(x) = (f \circ f \circ f \circ \dots \circ f)(x)$ ($n$ times), then $f_{32}(x) = $