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 $f$ be a composite function of $x$ defined by $f(u) = \frac{1}{u^2 + u - 2}$ and $u(x) = \frac{1}{x - 1}$. Then the number of points $x$ where $f$ is discontinuous is

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

The composite mapping $fog$ of the maps $f:R \to R$,$f(x) = \sin x$,and $g:R \to R$,$g(x) = x^2$ is:

If $f(x)=e^x$ and $g(x)=\ln(x)$ for all $x \in [1, \infty)$,then $f \circ g$ is . . . . . .

If $f(x) = \frac{1}{1 - x}$,then the derivative of the composite function $f[f\{ f(x)\} ]$ is equal to