## anonymous one year ago Determine the domain of a function, algebraically. There are a limited type of functions that exist. (Ex: a function can be quadratic, squared, etc.) So I was wondering if there are any rules that can help determine what the domain of a given function is without graphing. I know of only one rule, that all linear functions have the domain of all real numbers. Does that rule also apply to quadratic functions? Also how would one determine the domain of an function algebraically, with functions like: $q(w) = \dfrac{w + 4}{w^2 + 1}$

1. anonymous

Generally when you have fractions to find the domain we set the denominator as follow $denominator \neq 0$ so for your function we have $w^2 +1 \neq 0$ this will give your restrictions for the domain.

2. anonymous

In interval notation, would we write that as: $(0, \infty)$

3. anonymous

Your domain for this should be all real numbers, are you sure it's w^2+1?

4. anonymous

and not w^2-1

5. anonymous

I am quite sure it is, + 1, and not part of the exponent. Is there a reason you think it is, -1?

6. anonymous

Hmm ok, just wondering you would have all real numbers for this, but if you had $\frac{ w+4 }{ w^2-1 }$ then we would have the following $w^2 - 1 \neq 0 \implies w = \pm 1$ as our restrictions

7. anonymous

I see, thank you. So to recap, would: $u(x) = \dfrac{x - 5}{2x + 4}$ domain be all real numbers too?

8. anonymous

Not quite, since it's a fraction lets do the same as above to find the restrictions in the domain by setting it not equal to 0. $2x+4 \neq 0 \implies 2x \neq -4 \implies x =-\frac{ 4 }{ 2 } = -2$ so our domain is all real numbers except for x = -2

9. anonymous

Also for the above example, $\frac{ w+4 }{ w^2-1 }$ to put it in interval notation, it would be $(- \infty, -1) \cup (-1,1) \cup (1, \infty )$

10. anonymous

Ah! I understand now, so the domain has a restriction, at a value of $$x$$ when $$x$$ makes the denominator of a fraction equal zero. The the interval notation would be: $\Large{(-\infty, -2)\cup (-2, 2) \cup (2, \infty) }$ The values in the center of the notation, $$(-2, 2)$$ are the ones we avoid, correct?

11. zzr0ck3r

The general problem are 1) dividing by zero 2) taking the square root (or any even root) of a negative number 3) putting non positive numbers in a log function

12. anonymous

2 is in the domain, also note the first one is all real numbers, because otherwise we have complex numbers. $(- \infty, \infty)$ (first one)

13. anonymous

So for the second one, would the proper notation be this: $\Large{(-\infty, -2)\cup (-2, \infty) }$

14. zzr0ck3r

or $$\{x\mid x\ne 2\}$$

15. anonymous

Yup! Looks good! :)

16. anonymous

Note that if you have square brackets [...] that means "included" if it's open brackets (...) that means not including, that's why we have it as-2) <- open brackets

17. anonymous

Thank you, both for your help.

18. anonymous

Anytime :)