If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=x^{3}+5$,then $(fog)^{-1}(x) = $

For every real number $x \neq -1$, let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

For $x \in R$,two real-valued functions $f(x)$ and $g(x)$ are such that $g(x) = \sqrt{x} + 1$ and $(f \circ g)(x) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to:

Let $D = \mathbb{R} - \{0, 1\}$ and $f: D \rightarrow D$,$g: D \rightarrow D$,and $h: D \rightarrow D$ be three functions defined by $f(x) = \frac{1}{x}$,$g(x) = 1 - x$,and $h(x) = \frac{1}{1 - x}$. If $j: D \rightarrow D$ is such that $(g \circ j \circ f)(x) = f(x)$ for all $x \in D$,then which one of the following is $j(x)$?

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\}$ is equal to

If $f(x) = x^2 + 1$,then $fof(x)$ is equal to