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September 1, 2014

September 1, 2014

Posted by **Anonymous** on Friday, July 12, 2013 at 4:23pm.

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 -
**Steve**, Friday, July 12, 2013 at 4:30pmfirst 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

so,

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

= -|x| - (|x|)^2 + 1 ≠ f(|x|)

so it is not an even function of |x|.

- Math -
**Anonymous**, Friday, July 12, 2013 at 5:28pmNo it's written that way on my assignment.. thats why I'm confused.

- Math -
**Steve**, Friday, July 12, 2013 at 5:34pmme 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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