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anonymous

  • 5 years ago

X^4-10x^2+24< x^3-4x Write in interval notation, use bracket for single points

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  1. amistre64
    • 5 years ago
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    x^2 = -6,4 for the LHS if I did it right in me head. x = 0,2,-2 for the RHS..... I wonder if that helps....

  2. amistre64
    • 5 years ago
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    x^4 +x^3 -10x^2 +4x +24 < 0

  3. anonymous
    • 5 years ago
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    thanks

  4. amistre64
    • 5 years ago
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    that should be ...-x^3.... typo :)

  5. anonymous
    • 5 years ago
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    So, what goes inside the brackets

  6. amistre64
    • 5 years ago
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    dunno yet, still trying to factor it, I got so far: (x-2)(x+2)(x+2)(x-3) We put these on a number line to see which of these makes everything less than 0....

  7. amistre64
    • 5 years ago
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    if I did it right: (-inf,-2) or (2,3)

  8. amistre64
    • 5 years ago
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    the easiest way to determie the values are to draw a number line and put in the values that make the equation equal to zero. Take each value and to the left of it, put a (-) and to the right of it put a (+); then multiply your signs together to get area that are (+) and (-). then pick the areas that fit the equation :)

  9. anonymous
    • 5 years ago
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    I got x3+3x^2-4x-12

  10. anonymous
    • 5 years ago
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    Actually, once you have it factorised, it is probably easier to quickly sketch the graph of the quartic function.

  11. amistre64
    • 5 years ago
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    or do some calculus :)

  12. amistre64
    • 5 years ago
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    X^4-10x^2+24< x^3-4x subtract the RHS from the LHS. x^4 -x^3 -10x^2 +4x +24 >0 right?

  13. amistre64
    • 5 years ago
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    <0 .... typo again... its these fat fingers and this tiny little keyboard :)

  14. anonymous
    • 5 years ago
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    yes

  15. amistre64
    • 5 years ago
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    dont let the "hero" title fool ya, I am after all, just an idiot in disguise :)

  16. amistre64
    • 5 years ago
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    the LHS factors to: (x-2)(x+2)(x-3)(x+2) right? remultiply to check thatits right...

  17. anonymous
    • 5 years ago
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    M Dot cancelled a factor of (x+2) at some point... best not to ever do that with a factor unless you can be sure it is >0 ∀ x ∈ R or it flips the inequality.

  18. amistre64
    • 5 years ago
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    <.......-2.........2.........3.......> - + + + - - + + - - - + -------------------------- - + - + if I did it right, these are my results for the number line....

  19. amistre64
    • 5 years ago
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    but something seems off, the even degree should be positive at both ends...

  20. anonymous
    • 5 years ago
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    The answer is only 2< x <3 , not x<-2

  21. amistre64
    • 5 years ago
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    ahhh..... that seemes more plausible :)

  22. anonymous
    • 5 years ago
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    x^4-10x^2+24\le x^3-4x. The instructor wants x\[\epsilon\] and U

  23. amistre64
    • 5 years ago
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    U is the universal set right?

  24. anonymous
    • 5 years ago
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    right

  25. amistre64
    • 5 years ago
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    can you do calculus on it? :)

  26. anonymous
    • 5 years ago
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    This is an algebra class. I'm familiar with any calculus

  27. anonymous
    • 5 years ago
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    I'm not familiar

  28. amistre64
    • 5 years ago
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    bummer.... calculus would allow us to "see" what is happening at cetain points on the graph more easily

  29. amistre64
    • 5 years ago
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    it appears that the graph touches then turns at x = -2 .... which is why it is positive on both sides of it....

  30. amistre64
    • 5 years ago
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    so it "crosses" the x axis at 2 and 3 which make it go negative AND less than 0

  31. amistre64
    • 5 years ago
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    so your notation would be [x|x is an element of (2,3)} or some such notation

  32. anonymous
    • 5 years ago
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    x \[\in \cup\]

  33. amistre64
    • 5 years ago
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    I am unclear about the "U" set, its been qwhile since I had to play in set theory :)

  34. anonymous
    • 5 years ago
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    Ok, I'm going to go with my gut.

  35. amistre64
    • 5 years ago
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    gut going is good :)

  36. anonymous
    • 5 years ago
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    thank you for your help

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