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:

Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

Find the function $g(t)$ if $f(t)=3t-2$ and $(g \circ f)^{-1}(t)=t-2$.

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

Two functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined as follows: $f(x) = \begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} -1, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases}$. Then, $(f \circ g)(\pi) + (g \circ f)(e)$ is equal to:

Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).