even or odd function? h(x) = -x^3 / 3x^2 - 9

[racca]

I have to determine whether this function is even or odd.

h(x)=-x^3/3x^2-9

I know that when a function is odd f(-x)=-f(x) but how does it work when the variable is already negative?

Can someone please help me work through this problem step by step

 

[Loren]

Re: even or odd function?

h(x)=-x^3/3x^2-9 means \(\displaystyle h(x)=\frac{-x^3}{3x^2}-9\).

If you don't mean that, please use parenthesis to clarify.

 

[Deleted member 4993]

Re: even or odd function?

Suppose:

\(\displaystyle f(x) \, = \, - \, [x^3]\)

then

\(\displaystyle f(-x) \, = \, - \, [(-x)^3] \, = \, - \, [-x^3] \, \, = \, x^3\)

so then is f(x) an odd function or even function?

 

[stapel]

how does it work when the variable is already negative?

The variable is not "already negative"; it merely has a "minus" sign in front of one of its appearances. The variable is what it is, and its sign won't be determined until you plug a number in for the variable.

To see this a bit more clearly, try plugging numbers in. If x = 2, then -x[sup:3jq50lqf]3[/sup:3jq50lqf] = -(2)[sup:3jq50lqf]3[/sup:3jq50lqf] = -(8) = -8; but if x = -2, then -x[sup:3jq50lqf]3[/sup:3jq50lqf] = -(-2)[sup:3jq50lqf]3[/sup:3jq50lqf] = -(-8) = +8.

For the "even, odd, or neither" thing, you're putting a "minus" sign on the variable, and then plugging that into the expression. If you replace "x" with "-x", then -x[sup:3jq50lqf]3[/sup:3jq50lqf] becomes -(-x)[sup:3jq50lqf]3[/sup:3jq50lqf] = -[(-x)(-x)(-x)] = -[-x[sup:3jq50lqf]3[/sup:3jq50lqf]] = +x[sup:3jq50lqf]3[/sup:3jq50lqf].

Hope that helps!

Eliz.