Posted by Anonymous on .
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,
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,
No it's written that way on my assignment.. that's why I'm confused.

Math 
Steve,
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.