kitchen table math, the sequel: help desk - even and odd functions

Wednesday, April 20, 2011

help desk - even and odd functions

5. Determine whether each function is even, odd, or neither.
a. f(x) = x^4 = 3x^2 + 5
b. g(x) = 2x^2 + 6x + 4

Algebra and Trigonometry Structure and Method
Dolciani

Algebra and Trigonometry: Structure and Method, Book 2
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6 comments:

LSquared32 said...

You would like help, yes?

The first is even. Even means symmetric about the y-axis. All polynomials with only odd exponents are even: a constant is even, and x^even is even. The exponents in the first polynomial are 4, 2, 0--even function.

The second is neither. It is a polynomial with both even and odd exponents--certain to be neither odd nor even. To prove it, we can evaluate at 1 and -1: (1,12), (-1, 0). If it were symmetric about the y-axus, the y-coordinates would be the same for both. Not symmetric about the y-axis, thus not even.

Odd means symmetric around the origin. Polynomials with only odd exponents are odd: x^odd is an odd function. Both of these polynomials have constant terms, so they are not odd. To prove the second is not odd, use the same two points as last time (1,12), (-1,0). If it were symmetric about the origin, the y-coordinate would be (-1)* the y-coordinate of the first. Points (1,12) and (-1,-12) would be symmetric about the origin. (1,12) and (-1,0) are not, thus the second polynomial is neither.

You can use the shortcut of looking at the exponents for (finite) polynomials. For other functions (trig functions, cube roots, etc.), you need the symmetry definition to check if they are odd or even.

Does that fill in the gaps you were hoping for?

April 20, 2011 at 12:35 PM
ChemProf said...

I assume the first one should be x^4 + 3x^2 + 5?

This one is even because if you let x = -x, you get the same function (since (-x)^2 = x^2).

The other one is neither even or odd because if you let x = -x, then the new function is 2x^2 - 6x + 4, so is not identical to the original function nor is it the original function times negative 1.

April 20, 2011 at 12:35 PM
Catherine Johnson said...

thank you SO MUCH

we are back in math hell....

which I can't really post about

THANK YOU

fyi: we've hired a tutor & C. says that the tutor was unclear about this question (I don't know whether that's true --- C. may have the wrong impression)

the tutor is fantastic, btw

basically, we need a live-in math teacher

April 20, 2011 at 2:22 PM
rocky said...

As ChemProf said, an even function has the property that
f(-x) = f(x)

Similarly, an odd function has the property that
f(-x) = -f(x)

So the thing to do is plug in -x and try to manipulate it into either f(x) or -f(x)

WHAT IT MEANS
If a function is even then if you start at x = 0 (the y axis), then
f(-1) = f(1)
f(-2) = f(2) etc.
In other words, the function in "negative land" would be the exact mirror image of the function in "positive land". It is symmetric about the y axis. Every point (x, y) has a corresponding point (-x, y).

For an odd function
f(-1) = -f(1)
f(-2) = -f(2) etc.
In other words, the function in "negative land" is an upside-down mirror image of the function in "positive land". People call this "symmetric about the origin" (like a pinhole camera) because every point (x, y) has a corresponding point (-x, -y).

April 20, 2011 at 4:10 PM
tjb said...

And you can remember "odd" and "even" by thinking of the very simplest polynomials:

x^2, x^4, x^8, etc are even

x, x^3, x^5, etc are odd

April 20, 2011 at 5:20 PM
Michael Weiss said...

One of my favorite, simple theorems is: Any function can be decomposed into a sum of an odd function and an even function.

How? If f(x) is the given function, define e(x) = ( f(x) + f(-x) ) / 2, and o(x) = ( f(x) - f(-x) ) / 2. Verify that e(x) is even, o(x) is odd, and e(x) + o(x) = f(x).

So looking at any function you can talk about the "odd part" of the function and the "even part" of the function. In the case of a polynomial, this is especially simple: The "odd part" is the terms of odd degree, and the "even part" is the terms of even degree.

Extra credit: Prove that this decomposition is unique... in other words, that if e1(x) + o1(x) and e2(x) + o2(x) are two decompositions of f(x) into even and odd parts, then e1(x) = e2(x) and o1(x) = o2(x).

April 21, 2011 at 10:05 AM

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