finding domain of a function

[coolbeans33]

R(x)= ( x²-2x -8)/ (x² - 16). I get that x doesn't equal 4, but do I also have to set the numerator to zero? How come it's a problem if you divide by zero, but not if you divide zero by a number?

 

[srmichael]

R(x)= x²-2x -8/ x² - 16. I get that x doesn't equal 4, but do I also have to set the numerator to zero? How come it's a problem if you divide by zero, but not if you divide zero by a number?

I kind of like to think of the domain in laymen's terms as all the x values that you can plug into a function that doesn't "blow up" the function and by "blowing up" I mean, for example, division by 0, square root of a negative number, etc.

That being said, sometimes it is easiest to find the domain of x by finding the values of x not in the domain, that is, the values of x that would "blow up" the function. That is why you would set the denominator equal to 0 to find those values of x (if any) that would make the denominator 0. In this example, x² - 16 = 0 thus x² = 16 and thus x = ± 4.

So the domain in interval notation would be: (-∞, -4) U (-4, 4) U (4, ∞)

You only have to set the numerator equal to 0 if you are looking for the x intercepts as all x intercepts have the coordinates of (x, 0) and the only way that a rational function can equal 0 is if the numerator equals 0.

 

[mathmari]

R(x)= x²-2x -8/ x² - 16. I get that x doesn't equal 4, but do I also have to set the numerator to zero? How come it's a problem if you divide by zero, but not if you divide zero by a number?

To find the domain of the function \(\displaystyle R(x)=\frac{x^2-2x-8}{x^2-16} \), the denominator must be different to zero.

\(\displaystyle x^2-16 = 0 \Rightarrow x^2=16 \Rightarrow x=\pm 4 \)

So, the domain is \(\displaystyle \mathbb{R} \) -{ \(\displaystyle \pm 4 \) }.