Let $f(x)=x-\frac{1}{x}$, then $f(-1)$ is

Let $f, g: R \rightarrow R$ be two functions defined as $f(x)=|x|+x$ and $g(x)=|x|-x \forall x \in R$. Then $(f \circ g)(x)$ for $x<0$ is

A is a set having 6 distinct elements. The number of distinct functions from $A$ to $A$ which are not bijections is

Let $f: R \rightarrow R$ be defined by
$f(x)=\left\{\begin{array}{lc}2 x ; & x > 3 \\ x^2 ; & 1 < x \leq 3 . \text { Then } \\ 3 x ; & x \leq 1\end{array}\right.$

$$ f(-1)+f(2)+f(4) \text { is }$$