Final answer:
The function f(x) = x^{-4} + cos(-2x) is even because both x^{-4} and cos(-2x) exhibit even symmetry. This means f(-x) = f(x) for all x in the domain of the function.
Step-by-step explanation:
To determine whether the function f(x) = x^{-4} + \cos(-2x) is even, odd, or neither, we evaluate its symmetry properties. A function is even if f(-x) = f(x) for all x in the function's domain, and odd if f(-x) = -f(x) for all x in the domain.
First, consider x^{-4}. This part is an even function because raising any real number to a negative even power is the same as raising its opposite to that power: x^{-4} = 1/x^{4} and (-x)^{-4} = 1/(-x)^4, which is equal to 1/x^{4} since the negatives cancel out.
Next, consider \cos(-2x). The cosine function is an even function, and \cos(-\theta) = \cos(\theta). Consequently, \cos(-2x) = \cos(2x), preserving the even symmetry.
Since both parts of the function are even, the function f(x) = x^{-4} + \cos(-2x) is even.
Therefore, the correct answer is: a) Even.