Posted by Anonymous on Friday, July 12, 2013 at 4:23pm.
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  Steve, Friday, July 12, 2013 at 4:30pm
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
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 5:28pm
No it's written that way on my assignment.. that's why I'm confused.

Math  Steve, Friday, July 12, 2013 at 5:34pm
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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