[tex]f:D \rightarrow \mathbb_{R},$ $ \\ \\ f) $ $ f(x) =
\left\{ \begin{array}{ll} 4x+1,~~~ x\in[-1,0] \\ 1-4x, ~~~x \in(0,1]\end{array} \right \\ \\ f(-x) = \left\{ \begin{array}{ll} 4(-x)+1,~~~x\in[-1,0] \\ 1-4(-x), ~~~x \in(0,1]\end{array} \right= \\ \\ = \left\{ \begin{array}{ll} -4x+1,~~~ x\in[-1,0] \\ 1+4x, ~~x \in(0,1]\end{array} \right =\left\{ \begin{array}{ll} 1-4x,~~~ x\in[-1,0] \\ 4x+1, ~~x \in(0,1]\end{array} \right =\\ \\ \Rightarrow f(-x) = f(x) \Rightarrow f~ para.[/tex]
[tex] g)$ $f(x) = x-\sqrt{x^2+1} \\ f(-x) = -x-\sqrt{(-x)^2+1} = -x-\sqrt{x^2+1} \\ \\ $Functia nu este nici para nici impara.[/tex]
