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Is the function f(x) = +|-2 even, odd, or neither?

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Final answer:

The function f(x) = +|-2 is even because it satisfies the condition y(x) = y(-x), meaning the function is symmetric about the y-axis. It is not odd because it does not satisfy the condition y(x) = −y(-x).

Step-by-step explanation:

To determine if the function f(x) = +|-2 is even, odd, or neither, we first need to understand the definitions of even and odd functions:

  • An even function meets the condition y(x) = y(-x), which means the function is symmetric about the y-axis.
  • An odd function satisfies y(x) = −y(-x), meaning that the function is symmetric about the origin, and if you reflect it about the y-axis and then the x-axis, you will get the original function.

We can test the function f(x) by evaluating f(-x) and checking both conditions:

For f(-x):
Since f(x) is simply the constant +|-2 (which is just 2), f(-x) will also be +|-2. Therefore, we have:

  • f(x) = 2 for any
  • f(-x) = 2 for any x

As we can see:

  • f(x) = f(-x), which means the function is even.

It is not odd because an odd function would have to satisfy f(-x) = -f(x), and in this case, -f(x) would be -2, which is not equal to f(-x) = 2.

User Vladimir Mironov
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