PDA

View Full Version : odd vs even


UrbanXrisis
Oct6-04, 08:39 PM
the rule for an odd function is: -f(x)=f(x) correct?

however, x^3 is odd? Why is that? -(x^3) != (x^3)


Also, how would someone tell if a function is odd or even if it was an equation like: (x^7)(x^6)/(x^4) or something of that nature?

Gokul43201
Oct6-04, 08:44 PM
the rule for an odd function is: -f(x)=f(x) correct?

however, x^3 is odd? Why is that? -(x^3) != (x^3)


Also, how would someone tell if a function is odd or even if it was an equation like: (x^7)(x^6)/(x^4) or something of that nature?

You've got the definition wrong. You're essentially stating that -something = something. This is true only if something = 0.

Correct definition : If f(-x) = -f(x), then f is odd.

UrbanXrisis
Oct6-04, 08:45 PM
ahhh okay! That makes sense! What about telling if a function is odd or even if it was an equation like: (x^7)(x^6)/(x^4) or something similar to that?

Gokul43201
Oct6-04, 08:48 PM
ahhh okay! That makes sense! What about telling if a function is odd or even if it was an equation like: (x^7)(x^6)/(x^4) or something similar to that?

What you've written can be simplified to x^9. (since x^ax^b = x^{a+b})

Odd powers of a variable are odd functions. And even powers are even functions.

arildno
Oct6-04, 08:49 PM
Plug in -x at the x place. If what comes out of f(-x) is EXACTLY -f(x), then your function is odd.

UrbanXrisis
Oct6-04, 08:50 PM
what if a function was...[(x^7)+(x^6)]/(x^4)

Gokul43201
Oct6-04, 08:50 PM
f(x) = x^3 + x^2 is neither odd nor even, (this is what your example simplifies to). See why this is true, by applying the definition.

UrbanXrisis
Oct6-04, 09:00 PM
Because the exponet is an odd and even number? So it's neither. Does the sign make any difference? Positive or negative? what if a function was...[(x^7)+(x^6)]/[(x^4)-(x^3)]

Gokul43201
Oct6-04, 09:02 PM
Read post #5.

UrbanXrisis
Oct6-04, 09:06 PM
I get the point :smile: thanks

TSN79
Dec8-04, 12:36 PM
What if the function is defined differently at different intervals? How would I then go about finding out whether it's odd or even?