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roselin

explain why the equation cox=x has at least one solution.

  • one year ago
  • one year ago

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  1. zepdrix
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    cox???

    • one year ago
  2. roselin
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    sorry,sorry. its cos x

    • one year ago
  3. zepdrix
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    oh lol ^^

    • one year ago
  4. zepdrix
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    Hmm do you remember what the graph for cosine looks like?

    • one year ago
  5. roselin
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    somewhat

    • one year ago
  6. baldymcgee6
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    I'm pretty sure cos(x) = x only has one solution

    • one year ago
  7. baldymcgee6
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    http://screencast.com/t/sLkE6Ylvzpk

    • one year ago
  8. zepdrix
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    |dw:1351373928927:dw| This is kind of the idea rose :D At this particular point, the x value and the cosx value are the same.

    • one year ago
  9. roselin
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    |dw:1351373944681:dw|

    • one year ago
  10. zepdrix
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    So in your picture, x=pi/2 but cosx = 0 Hmm that point doesn't work so well :c

    • one year ago
  11. roselin
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    okay, i will consider the picture that you showed

    • one year ago
  12. zepdrix
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    Yah I can't think of the proper way to explain it :D It probably has something to do with the fact that cosine oscillates back and forth between 1 and -1, and since it is continuous it would have to have a solution :d i dunno... just think about it i guess :3 heh

    • one year ago
  13. baldymcgee6
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    Since the function y = x has a domain of all real numbers... as is cos(x), there must be a point of intersection somewhere

    • one year ago
  14. roselin
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    okay,

    • one year ago
  15. roselin
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    So which one should I consider now? i have two answers here

    • one year ago
  16. baldymcgee6
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    Combine them... you could say.. Since both y=x and the cosine function have a domain of all real numbers and both are continuous at least one point of intersection is definite.

    • one year ago
  17. roselin
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    oh okay, thank you so much guys

    • one year ago
  18. Zarkon
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    use the intermediate value theorem

    • one year ago
  19. latremese40
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    roelin i think you did it right

    • one year ago
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