## warpedkitten one year ago If f(x) = x^2 - 25 and g(x) = x - 5, what is the domain of (f/g)(x)?

1. UsukiDoll

$f(x) = x^2-25$ and $g(x) = x-5$ $(\frac{f}{g})(x)$ is this your question and are those the right functions?

2. warpedkitten

Yes, that's correct.

3. UsukiDoll

alright so we need $(\frac{f}{g})(x)$ $(\frac{f}{g})(x) = \frac{x^2-25}{x-5}$ can you factor the $x^2-25$? after that step we can cancel a term out

4. warpedkitten

I don't know how to solve it, I'm not very good with math. lol.

5. UsukiDoll

oh ok. no problem . we can use the difference of squares formula $(a^2-b^2) = (a+b)(a-b)$ so if we let a = x and b = 5 $(x^2-5^2) = (x+5)(x-5)$

6. UsukiDoll

so $(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}$ so what does the numerator and denominator have in common

7. UsukiDoll

there's a x-5 in the numerator and denominator so that term is canceled out

8. UsukiDoll

$(\frac{f}{g})(x) = x+5$ since our new function isn't a fraction , there are no restrictions so the domain is all real numbers.

9. UsukiDoll

the graph should be a straight line and it's a function because it passes the vertical line test. A vertical line test is needed to determine if a graph is a function or not. If it's a function then the vertical line should hit the graph only once. If it's not a function, the vertical line crosses the graph more than once.

10. warpedkitten

So my answer options are all real values of x all real values of x except x = 5 all real values of x except x = –5 all real values of x except x = 5 and x = –5 Would that mean it's a? Since they're all real numbers? Or would it be b?

11. UsukiDoll

has to be a... all reals. the function is not a fraction

12. warpedkitten

Thank you! :)

13. UsukiDoll

this is the graph of the function (f/g) (x)

14. UsukiDoll

if you draw a vertical line on this graph, the vertical line only touches once, so it's a function .

15. zzr0ck3r

hehe domain does not include x=5. It would not show up on a graph because its a hole. If you graph $$\frac{x^2-25}{(x-5)}$$ and zoomed in infinitely close it would look like this |dw:1439266602373:dw| $\frac{x^2-25}{(x-5)}\ne x+5$ in general (one is defined for 5, one is not) But if we remove $$5$$ from the domain it is :)

16. UsukiDoll

if we factor the numerator, the x-5 cancels out leaving us with the new function no longer being a fraction. Hence, no restrictions... all reals. $(\frac{f}{g})(x) = \frac{x^2-25}{x-5}$ $(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}$ $(\frac{f}{g})(x) = x+5$

17. UsukiDoll

if you're not factoring then there is a restriction which is all reals except when x = 5. But I think that there have been times when canceling terms can happen.

18. UsukiDoll

both $(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}$ and $(\frac{f}{g})(x) = x+5$ produce different results.. It's like do we consider the factored version in that case it's all reals on the domain or the non-factored case all reals except x = 5 ?