Question #5785b

Jun 5, 2017

$f \ne g .$

Explanation:

Let us first find out all the Images of

$x \in D \text{ under } f , i . e , f \left(x\right) , \forall x \in D .$

$f \left(- 2\right) = | - 2 | + 2 \left(- 2\right) = 2 - 4 = - 2.$

Similarly, $f \left(- 1\right) = - 1 , f \left(0\right) = 0 , f \left(1\right) = 3 , f \left(2\right) = 6.$

Therefore, if we represent $f$ as the Set of Ordered Pairs, then,

we have, $f = \left\{\begin{matrix}- 2 & - 2 \\ - 1 & - 1 \\ 0 & 0 \\ 1 & 3 \\ 2 & 6\end{matrix}\right\} .$

Similarly, $g = \left\{\begin{matrix}- 2 & 6 \\ - 1 & 3 \\ 0 & 0 \\ 1 & - 1 \\ 2 & - 2\end{matrix}\right\} .$

Clearly, $f \ne g .$

Enjoy Maths.!