Question

How do you determine if `f(x)=sinxsqrt(x²+1)` is an even or odd function?

Answer 1

`f(x)` is an odd function.

Explanation:

If `f(-x)=f(x)`, it is an even function.
If `f(-x)=-f(x)`, it is an odd function.
If neither equation is true, the function is neither even nor odd.

`f(-x)`
`=sin(-x)sqrt((-x)^2+1)`
`=-sin(x)sqrt((-x)^2+1)`
`=-sin(x)sqrt(x^2+1)`
`=-f(x)`

Thus, `f(x)` is an odd function.

Now, what does it mean by a function being even or odd?

Well, if a function is even, it is symmetric to the `y` axis.
Well, if a function is odd, it is symmetric to the origin.

What if a function were to be symmetric to the `x` axis?
An example for this would be `y=+-sqrtx`. This is indeed symmetric to the `x` axis, but it has two `y` values for a given `x` value. This means that it is not a function.