Question

Determine whether the function is even, odd, or neither

h(x)=x^3-5

Answer 1

\[h(x)=x^3-5\]

Answer 2

3 is odd and it slides down the y which messes it all up i believe

Answer 3

whenever we move an odd function away from the origin; it loses its self

Answer 4

whenever we shift an even function away from the y axis, it loses is self

Answer 5

\(f(-x)=-x^3-5=-(x^3+5)\ne f(x) \text{ and } f(-x)\ne-f(x)\). So, it's neither.

Answer 6

yes ^ it is neither he is correct

Answer 7

it should be \(h(x)\), instead of \(f(x)\).

Answer 8

Let \[h(x)=x^3-5.\] For the case in which x is even, x^3 is also even, and an even number added or subtracted from an odd number is always an odd number. Inversely, for the case in which x is odd, x^3 is also odd, and an odd number added or subtracted from an odd number is always an even number. Therefore, the function is neither even nor odd.

Answer 9

I don't think this has anything to do with the number being odd or even. Even and odd functions are functions that has a certain relation of symmetry. It would give the same result if \(h(x)=x^3-4\), or any other even number for that matter.

Answer 10

I'm not sure what the definition is, but if you change the function to
\[h(x)=x^3-4,\]the conclusion of the above "proof" will be the same regardless.

Answer 11

A function \(f\) is called an even function if \(f(x)=f(-x)\), and called odd if \(f(x)=-f(x)\). What I meant is that h(x)=x^3-4 is still neither odd nor even, but your proof would give a different result. That's if x is odd then x^3 is also odd and adding odd to an even number gives an odd number.

Answer 12

... and if x is even, then x^3 is also even and an even number plus an even number is an even number. Thus it's still neither odd nor even. But I won't keep pressing the argument.

Answer 13

You should conclude that it's both even and odd then!!