Question

Is `f(x)=x^2+sin x` an even or odd function?

Answer 1

It is neither.

Explanation:

A function `f` is even if and only if `f(-x)=f(x)` for every `x` in the domain of `f`.
A function `f` is odd if and only if `f(-x)=-f(x)` for every `x` in the domain of `f`.

For `f(x)=x^2+sin x`, we check `f(-x)`.

If `f(-x)` simplifies to `f(x)`, then `f` is even,
If `f(-x)` simplifies to `-f(x) =-(x^2+sin x)`, the `f` is odd.
If `f(-x)` does not simplify to one of the above, then `f` is neither even nor odd.

`f(-x) = (-x)^2+sin(-x)`

`= x^2-sinx` which is neither `f(x)` nor `-f(x)`

We cannot show that `f(-x)=f(x)` for every `x` by showing that it is true for only one `x`,
but we can show that is fails to be true for all `x` by showing that it fails for one value.

`f(pi/2)=pi^2/4+1`
`f(-pi/2) =pi^2/4-1`

It is, I think, clear that these numbers are neither equal nor negatives of each other.

Answer 2

Neither.

`f(-pi/6) = pi^2/36 - 1/2 ~~ 10/36 - 1/2 = -2/9`

`f(pi/6) = pi^2/36 + 1/2 ~~ 10/36 + 1/2 = 7/9`

So `f(-pi/6) != +-f(pi/6)`

Explanation:

An even function requires `f(-x) = f(x)`

An odd function requires `f(-x) = -f(x)`

Our example satisfies neither of these conditions.