Determine whether the function f(x)=|x|+x2+0.001 is even , odd or neither. 10 points

ANSWER
even
EXPLANATION
The given function is
[tex]f(x) = |x| + {x}^{2} + 0.001[/tex]
If f(x) is even then f(-x) =f(x).
[tex]f( - x) = | - x| + {( - x)}^{2} + 0.001[/tex]
[tex]f( - x) = | x| + {x}^{2} + 0.001[/tex]
We can see that:
[tex]f( - x) = f(x)[/tex]
Hence the given function is even.