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

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

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