# Characteristics of Polynomials

• January 3rd 2009, 09:04 AM
magentarita
Characteristics of Polynomials
Determine which of the following polynomial functions is even, odd or neither.

(a) f(x) = 3x^4 - 2x + 1...The answer is neither. BUT what makes it neither?

(b) f(x) = 3x^9 - 2x^5 + 1...The answer is neither. BUT what makes it neither?

Does the constant 1 make the two functions above neither?

Even functions have even powers and odd functions have odd powers. Am I right?

Neither functions involve functions that have both even and odd powers. Am I right?

I am still unclear about the "neither" concept.

To me, function (a) is even and and function (b) is odd.

The textbook tells me that I am wrong.

Can someone explain in simple terms this "neither" concept?

I want to fully grasp all basic concepts of precalculus before starting calculus 1 next semester.

By the way, I passed precalculus last semester with an A minus thanks to the wonderful math teachers on this website.

• January 3rd 2009, 09:18 AM
Mush
Quote:

Originally Posted by magentarita
Determine which of the following polynomial functions is even, odd or neither.

(a) f(x) = 3x^4 - 2x + 1...The answer is neither. BUT what makes it neither?

(b) f(x) = 3x^9 - 2x^5 + 1...The answer is neither. BUT what makes it neither?

Does the constant 1 make the two functions above neither?

Even functions have even powers and odd functions have odd powers. Am I right?

Neither functions involve functions that have both even and odd powers. Am I right?

I am still unclear about the "neither" concept.

To me, function (a) is even and and function (b) is odd.

The textbook tells me that I am wrong.

Can someone explain in simple terms this "neither" concept?

I want to fully grasp all basic concepts of precalculus before starting calculus 1 next semester.

By the way, I passed precalculus last semester with an A minus thanks to the wonderful math teachers on this website.

An odd function is defined by $f(x)=-f(-x)$. An even function is defined by $f(x) =f(-x)$
Since $f(x) \neq f(-x)$ and $f(x) \neq -f(-x)$ the function is neither odd nor even...

A function is odd if it fulfills the condition $f(x)=-f(-x)$, and even if it fulfils $f(x) =f(-x)$. If it fulfills neither condition, then it is neither.

In other words. If you have a graph of an odd function, then if you pick ANY value of x, and note the value of f(x). Then go to the -x on the same graph, you should find that the value is negative the value of the f(x) you noted earlier. (see sin(x))

And for the graph of an even function, if you pick ANY value of x, and note the value of y=f(x), then go to -x on the same graph, you should find that the y value is the same as it was for positive x. (see cos(x))

For a function which is neither, you will not be able to do this for all values of x.
• January 3rd 2009, 04:13 PM
DaveDammit
Also, if not too complex in length I think you can also plug the equation into a graphing calculator, or graph it by hand and determine whether it is symmetrical to the origin or whether it contains y-axis symmetry to determine even, odd, or neither.
• January 4th 2009, 01:32 PM
magentarita
thanks but...
Quote:

Originally Posted by DaveDammit
Also, if not too complex in length I think you can also plug the equation into a graphing calculator, or graph it by hand and determine whether it is symmetrical to the origin or whether it contains y-axis symmetry to determine even, odd, or neither.

I like calculators but I enjoy solving by hand more than letting a machine do the work for me.
• January 4th 2009, 01:33 PM
magentarita
ok...Very informative
Quote:

Originally Posted by Mush
An odd function is defined by $f(x)=-f(-x)$. An even function is defined by $f(x) =f(-x)$
Since $f(x) \neq f(-x)$ and $f(x) \neq -f(-x)$ the function is neither odd nor even...

A function is odd if it fulfills the condition $f(x)=-f(-x)$, and even if it fulfils $f(x) =f(-x)$. If it fulfills neither condition, then it is neither.

In other words. If you have a graph of an odd function, then if you pick ANY value of x, and note the value of f(x). Then go to the -x on the same graph, you should find that the value is negative the value of the f(x) you noted earlier. (see sin(x))

And for the graph of an even function, if you pick ANY value of x, and note the value of y=f(x), then go to -x on the same graph, you should find that the y value is the same as it was for positive x. (see cos(x))

For a function which is neither, you will not be able to do this for all values of x.

Thank you for the infomation.
• January 5th 2009, 11:51 AM
HallsofIvy
The reason for the names "even" and "odd" functions is that, for polynomials like these, the polynomial is an "even" function if and only if all the exponents of x are even, an "odd" function if and only if all the exponents are even. If there are both even and odd exponents, the polynomial is neither even nor odd.

Here, $3x^4- 2x+ 1= 3x^4- 2x^1+ x^0$ has exponents 4, 1 and 0. 4 and 0 are even but 1 is odd so this is neither even nor odd.

Similarly, $3x^9 - 2x^5 + 1$ has exponents 9, 5 and 0. Here is IS the "1" that causes it to be "neither" because all the other exponents are odd.

(I had been wondering what in the world course you were taking that had such varied questions! That was a heck of a Pre-Calculus course.)
• January 5th 2009, 11:06 PM
magentarita
ok....
Quote:

Originally Posted by HallsofIvy
The reason for the names "even" and "odd" functions is that, for polynomials like these, the polynomial is an "even" function if and only if all the exponents of x are even, an "odd" function if and only if all the exponents are even. If there are both even and odd exponents, the polynomial is neither even nor odd.

Here, $3x^4- 2x+ 1= 3x^4- 2x^1+ x^0$ has exponents 4, 1 and 0. 4 and 0 are even but 1 is odd so this is neither even nor odd.

Similarly, $3x^9 - 2x^5 + 1$ has exponents 9, 5 and 0. Here is IS the "1" that causes it to be "neither" because all the other exponents are odd.

(I had been wondering what in the world course you were taking that had such varied questions! That was a heck of a Pre-Calculus course.)

I will post more questions on this interesting topic.