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Show work to determine if the relation is even, odd, or neither.

f(|x|) = |x|- x^2 + 1

I thought it was even because the absolute value and squared will cancel the negatives, but I don't know how to show this through math..

  • Math -

    first off, I think it should be written

    f(x) = |x| - x^2 + 1

    To prove it's even, recall that even means f(-x) = f(x). So,
    f(-x) = |-x| - (-x)^2 + 1 = |x - x^2 + 1 = f(x)
    so f(x) is even.

    Now, if you truly meant

    f(|x|) = |x| - x^2 + 1

    then you need to express x^2 in terms of |x|, which is (|x|)^2


    f(-|x|) = -|x| - (-|x|)^2 + 1
    = -|x| - (|x|)^2 + 1 ≠ f(|x|)
    so it is not an even function of |x|.

  • Math -

    No it's written that way on my assignment.. that's why I'm confused.

  • Math -

    me too. It's certainly an unorthodox way to define a relationship. It's like saying

    f(x^2+1) = x^3 - 2x + 7

    you'd have to redefine the polynomial in x into an expression in terms of (x^2+1). Possible, but rather obfuscatory.

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