Question

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

Answer 1

the answer is even is the right answer

Answer 2

even

Step-by-step explanation

The given function is

`f(x) = |x| + {x}^(2) + 0.001`

If f(x) is even then f(-x) =f(x).

`f( - x) = | - x| + {( - x)}^(2) + 0.001`

`f( - x) = | x| + {x}^(2) + 0.001`

We can see that:

`f( - x) = f(x)`

Hence the given function is even.