Question

How do you determine if `f(x) = sqrt(4x^5 + 7x^3)` is an even or odd function?

Answer 1

`f(x)` is neither even nor odd, whether we consider it as a Real valued or Complex valued function.

Explanation:

An even function is one for which `f(-x) = f(x)` for all `x` in the domain.

An odd function is one for which `f(-x) = -f(x)` for all `x` in the domain.

`color(white)()`
Given:

`f(x) = sqrt(4x^5+7x^3)`

Note that:

`x = 1` is in the (implicit) domain of `f(x)` and

`f(1) = sqrt(4+7) = sqrt(11)`

So:

• If `f(x)` is an even function then `x = -1` is also in the domain and `f(-1) = f(1)`

• If `f(x)` is an odd function then `x = -1` is also in the domain and `f(-1) = -f(1)`

But we find:

`4(-1)^5+7(-1)^3 = -4-7 = -11`

So if we are considering `f(x)` as a Real valued function, then:

• `f(-1) = sqrt(-11)` is not defined
• So `x=-1` is not part of the domain of `f(x)`
• So `f(x)` is neither even nor odd.

Alternatively, if we are considering `f(x)` as a Complex valued function, then:

• `f(-1) = sqrt(-11) = sqrt(11)i`
• So `x = -1` is part of the domain of `f(x)`
• `f(-1) != f(1)` and `f(-1) != -f(1)`
• So `f(x)` is neither even nor odd.

In either case we find that `f(x)` is neither even nor odd.