I am having trouble finding the domain of

x^3 + 5x

------------

x^2 - 6x + 8

Thanks for your Help

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- Nov 1st 2006, 02:37 PMBarret52[SOLVED] Finding the Domain of a quadratic
I am having trouble finding the domain of

x^3 + 5x

------------

x^2 - 6x + 8

Thanks for your Help - Nov 1st 2006, 04:24 PMtopsquark
Limitations on the domain of a function take the form of zeros in the denominator, negative numbers under a square root sign, and the like. Here we have a fraction so we want to look at where the denominator goes to zero.

$\displaystyle x^2 - 6x + 8 = (x - 4)(x - 2) = 0$

So the denominator is zero at x = 2 and x = 4. Thus these values for x cannot be in the domain. This is our only restriction, so the domain will be:

$\displaystyle ( -\infty, 2) \cup (2, 4) \cup (4, \infty)$

or, more simply, "all real x except 2 and 4."

-Dan