Domain and polynomial degree

Mar 2017
358
3
Massachusetts
Hi,

I hope someone can help. So I have the following function h(x) = \(\displaystyle f(x)\cdot\) (1/f(x)).

I'm trying to understand if the degree in the function changes why the domain would be different.

For example, let's say that f(x) = \(\displaystyle (x^2)\)-25...the domain for h(x) would be all real numbers such that x doesn't equal 5 or -5.

But what would the domain be if it was to the power of 3 instead of 2 (e.g. f(x) = \(\displaystyle (x^3)\)-25)? I would think that the domain would be all real numbers such that x doesn't not equal 5^(2/3), but I don't see this being represented as a hole when I use graphing technology - this makes me skeptical whether my thinking is right about this domain which has a power of 3.

Please let me know.

- Olivia
 
Last edited:

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
Hi,

I hope someone can help. So I have the following function h(x) = \(\displaystyle f(x)\cdot\) (1/f(x)).

I'm trying to understand if the degree in the function changes why the domain would be different.

For example, let's say that f(x) = \(\displaystyle x^2\)-25...the domain for h(x) would be x \(\displaystyle x \neg\) 5 and \(\displaystyle x \neg\) -5.

But what would the domain be if it was to the power of 3 instead of 2? I would think that the domain would be x \(\displaystyle \neg 5^(2/3)\), but I don't see this being represented as a hole when I use graphing technology - this makes me skeptical whether my thinking is right about this domain which has a power of 3.

Please let me know.

- Olivia
You are correct. The denominator cannot be zero. So, you cannot have $x^3-25=0$ or $x^3=25$. That occurs at $x=\sqrt[3]{25} = 5^{2/3}$. So, the domain would be $x \neq 5^{2/3}$
 
Mar 2017
358
3
Massachusetts
Okay sounds good. It also said in my textbook that if 'f' is of an even degree, there may be no values excluded from the domain... is this true? And if so, what is an example of this?
 
Mar 2017
358
3
Massachusetts
Never mind, I understand now what I was confused about. All good. Thanks for your help.