If $g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$,then one such function $f(x)=$

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

If $f(x) = \frac{2x + 1}{3x - 2}$,then $(fof)(2)$ is equal to

$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

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.

If $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$,then $f(f(x))$ is