anonymous one year ago plz help fan medal and testimonial which of the following are continuus for all real values of x? I. f(x)= x^2+5/x^2-1 ii. g(X)= 3/x^2+1 iii. h(X)={x-1} choices- ii and iii only I and ii only I only ii only

1. anonymous

I know for a fact that choice b is wrong.

2. anonymous

@dan815 @Math2400 @Camila1315 @notyourdroid

3. myininaya

what does { } mean?

4. myininaya

for the two fractions you can find where they are not continuous by finding when the denominator is zero

5. anonymous

parenthesis

6. myininaya

so h(x)=x-1?

7. anonymous

yes

8. myininaya

h(x)=x-1 is just a line then

9. myininaya

do you have any questions?

10. anonymous

so my suspicion was right lol h(X)=x-1 is my answer? lol

11. myininaya

I didn't say that

12. myininaya

did you find when the bottoms were zero (if any values) for the first two?

13. anonymous

there continuous but I put I and ii as my answer before and I was told it was wrong.

14. myininaya

I assume the first fraction is: $f(x)=\frac{x^2+5}{x^2-1} \\ \text{ can you solve } x^2-1=0 \\ \text{ the second fraction is } g(x)=\frac{3}{x^2+1} \\ \text{ can you solve } x^2+1=0$

15. anonymous

x=1 for both

16. myininaya

x^2-1=0 x^2=1 x=1 or x=-1 right? but how is 1^2+1 zero?

17. myininaya

so f is discontinuous at x=1 and also x=-1 but what do you notice about x^2+1=0?

18. anonymous

same as the first.

19. myininaya

no (1)^2+1 is not zero (-1)^2+1 is not zero

20. myininaya

both (1)^2+1 and (-1)^2+1 is 1+1 which is 2

21. myininaya

if I write x^2+1=0 as x^2=-1 If i try to think of a positive number to plug in...I'm gonna have to square it but you know a positive times a positive =? or if i try to think of a negative number to plug in...I'm gonna have to square it but you know a negative times a negative=? or if I plug in 0 which still isn't gonna work because 0^2 is not -1

22. anonymous

okay...

23. myininaya

I'm asking you to tell me what a positive times a positive will give and what a negative times a negative will give

24. ganeshie8

{x} usually refers to the fractional part of x

25. anonymous

+ x + = + - x - = +

26. myininaya

right x^2 is always positive or 0 -1 is negative there is no way x^2 can even be -1 over the real numbers

27. myininaya

there is no real solution to x^2+1=0

28. myininaya

therefore g=3/(x^2+1) has no discontinuities ok so h(x)=x-1 or h(x)={x-1} @magy33 ?

29. myininaya

I'm asking since @ganeshie8 pointed out { } means fractional part

30. anonymous

h(x)= x-1 but does the {} really have anything to do with the equation itself? I askto be sure for next time :)

31. myininaya

If it is there, it does

32. ganeshie8
33. anonymous

ok. I think then that from my understanding the answer is ii and iii

34. ganeshie8

I'm inclining toward ii only

35. myininaya

me to if h={x-1} and not x-1

36. ganeshie8

the fractional function h(x) = {x-1} is discontinuous at every integer

37. anonymous

lol I was half right. thnks for the help guys. lol I still have more questions to go before im done lol I see if I can answer if not ill post lol