Let $f(x) = \frac{x-1}{x+1}$,$x \in R - \{-1, 0, 1\}$. If $f^{n+1}(x) = f(f^n(x))$ for all $n \in N$,then $f^6(6) + f^7(7) = $

Let $N$ denote the set of all natural numbers,and $Z$ denote the set of all integers. Consider the functions $f: N \rightarrow Z$ and $g: Z \rightarrow N$ defined by $f(n) = \begin{cases} (n+1)/2 & \text{if } n \text{ is odd} \\ (4-n)/2 & \text{if } n \text{ is even} \end{cases}$ and $g(n) = \begin{cases} 3+2n & \text{if } n \geq 0 \\ -2n & \text{if } n < 0 \end{cases}$. Define $(g \circ f)(n) = g(f(n))$ for all $n \in N$,and $(f \circ g)(n) = f(g(n))$ for all $n \in Z$. Then which of the following statements is (are) True?

If $f(x) = (p - x^n)^{1/n}$, $p > 0$ and $n$ is a positive integer, then $f[f(x)]$ is equal to

Let $f: R \rightarrow R$ be defined as $f(x) = 2x - 1$ and $g: R - \{1\} \rightarrow R$ be defined as $g(x) = \frac{x - 1/2}{x - 1}$. Then the composition function $f(g(x))$ is:

The graph of $y = f(x)$ is shown. The number of solutions of the equation $f(f(x)) = 2$ is:

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\begin{cases} \log _e x & , x>0 \\ e^{-x} & , x \leq 0 \end{cases}$ and $g(x)=\begin{cases} x & , x \geq 0 \\ e^{x} & , x < 0 \end{cases}$. Then $gof: R \to R$ is . . . .