If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3x-2$ and $g(x)=x^2+2$, then $[(g \circ f)+(f \circ g)](x) = $

If $f(x) = \frac{4x+7}{7x-4}$,then the value of $f\{f[f(2)]\} =$ ?

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

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:

Show that if $f: R - \{\frac{7}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+4}{5x-7}$ and $g: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{7}{5}\}$ is defined by $g(x) = \frac{7x+4}{5x-3}$,then $f \circ g = I_{A}$ and $g \circ f = I_{B}$,where $A = R - \{\frac{3}{5}\}$,$B = R - \{\frac{7}{5}\}$; $I_{A}(x) = x, \forall x \in A$,$I_{B}(x) = x, \forall x \in B$ are called identity functions on sets $A$ and $B$,respectively.

Let $f$ and $g$ be two functions defined by $f(x) = \begin{cases} x+1, & x < 0 \\ |x-1|, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ 1, & x \geq 0 \end{cases}$. Then $(g \circ f)(x)$ is