Let $f$ and $g$ be functions defined by $f(x) = \frac{x}{x + 1}$ and $g(x) = \frac{x}{1 - x}$. Then $(fog)(x)$ is

Let $f(x) = ax + b$ and $g(x) = cx + d$. The condition $f(g(x)) = g(f(x))$ holds for all $x$ if and only if ...

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

If $f: R \rightarrow R$ is defined by $f(x) = \frac{x}{x^{2}+1}$,find $f(f(2))$.

If $f(x) = e^{2x}$ and $g(x) = \log \sqrt{x}$ $(x > 0)$,then $fog(x)$ is equal to

Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying $(f \circ g \circ g \circ f)(x)=(g \circ g \circ f)(x)$,where $(f \circ g)(x)=f(g(x))$,is