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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

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  1. anonymous
    • one year ago
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    I know for a fact that choice b is wrong.

  2. anonymous
    • one year ago
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    @dan815 @Math2400 @Camila1315 @notyourdroid

  3. myininaya
    • one year ago
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    what does { } mean?

  4. myininaya
    • one year ago
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    for the two fractions you can find where they are not continuous by finding when the denominator is zero

  5. anonymous
    • one year ago
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    parenthesis

  6. myininaya
    • one year ago
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    so h(x)=x-1?

  7. anonymous
    • one year ago
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    yes

  8. myininaya
    • one year ago
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    h(x)=x-1 is just a line then

  9. myininaya
    • one year ago
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    do you have any questions?

  10. anonymous
    • one year ago
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    so my suspicion was right lol h(X)=x-1 is my answer? lol

  11. myininaya
    • one year ago
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    I didn't say that

  12. myininaya
    • one year ago
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    did you find when the bottoms were zero (if any values) for the first two?

  13. anonymous
    • one year ago
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    there continuous but I put I and ii as my answer before and I was told it was wrong.

  14. myininaya
    • one year ago
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    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
    • one year ago
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    x=1 for both

  16. myininaya
    • one year ago
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    x^2-1=0 x^2=1 x=1 or x=-1 right? but how is 1^2+1 zero?

  17. myininaya
    • one year ago
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    so f is discontinuous at x=1 and also x=-1 but what do you notice about x^2+1=0?

  18. anonymous
    • one year ago
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    same as the first.

  19. myininaya
    • one year ago
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    no (1)^2+1 is not zero (-1)^2+1 is not zero

  20. myininaya
    • one year ago
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    both (1)^2+1 and (-1)^2+1 is 1+1 which is 2

  21. myininaya
    • one year ago
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    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
    • one year ago
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    okay...

  23. myininaya
    • one year ago
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    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
    • one year ago
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    {x} usually refers to the fractional part of x

  25. anonymous
    • one year ago
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    + x + = + - x - = +

  26. myininaya
    • one year ago
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    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
    • one year ago
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    there is no real solution to x^2+1=0

  28. myininaya
    • one year ago
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    therefore g=3/(x^2+1) has no discontinuities ok so h(x)=x-1 or h(x)={x-1} @magy33 ?

  29. myininaya
    • one year ago
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    I'm asking since @ganeshie8 pointed out { } means fractional part

  30. anonymous
    • one year ago
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    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
    • one year ago
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    If it is there, it does

  32. ganeshie8
    • one year ago
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    http://mathworld.wolfram.com/FractionalPart.html

  33. anonymous
    • one year ago
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    ok. I think then that from my understanding the answer is ii and iii

  34. ganeshie8
    • one year ago
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    I'm inclining toward ii only

  35. myininaya
    • one year ago
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    me to if h={x-1} and not x-1

  36. ganeshie8
    • one year ago
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    the fractional function `h(x) = {x-1}` is discontinuous at every integer

  37. anonymous
    • one year ago
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    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

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