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VIbarguen1

  • 2 years ago

Cal help? find the intervals for which f is increasing or decreasing and locate all relative extrema on (0,2pi). f(x)= cos x / 1 + sin ^ 2x

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  1. satellite73
    • 2 years ago
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    did you take the derivative?

  2. VIbarguen1
    • 2 years ago
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    Yep I got f'(x) = 2 sin(x) cos (2x) + 5 /cos (2x)-3 )^2

  3. VIbarguen1
    • 2 years ago
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    it is so long im not sure its right

  4. VIbarguen1
    • 2 years ago
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    @satellite73

  5. VIbarguen1
    • 2 years ago
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    @RaphaelFilgueiras

  6. RaphaelFilgueiras
    • 2 years ago
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    |dw:1389408369353:dw|

  7. VIbarguen1
    • 2 years ago
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    could you help me solve that I have no idea how

  8. RaphaelFilgueiras
    • 2 years ago
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    use sin²(x)+cos²(x)=1-> cos²(x)=1-sin²(x), and then you will have only sin x in your equations

  9. RaphaelFilgueiras
    • 2 years ago
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    |dw:1389408745656:dw|

  10. VIbarguen1
    • 2 years ago
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    raphael i'm sorry i am not following, where did the Z come from? and how does that get me to the extrema?

  11. RaphaelFilgueiras
    • 2 years ago
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    Z means k belong to the interger set,it's just a generical ans

  12. RaphaelFilgueiras
    • 2 years ago
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    x= 0 and 2 pi will be your solutions

  13. RaphaelFilgueiras
    • 2 years ago
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    and pi

  14. VIbarguen1
    • 2 years ago
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    is that my min and max solutions?

  15. VIbarguen1
    • 2 years ago
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    and it would be decreasing because its positive right?

  16. RaphaelFilgueiras
    • 2 years ago
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    min x= pi max are both x=0 and 2pi

  17. VIbarguen1
    • 2 years ago
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    ok got it

  18. RaphaelFilgueiras
    • 2 years ago
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    when sin(x)> 0 it will be decreasing x in]0,pi[ when sinx(x)<0 it will be increasing x in ]pi,2pi[

  19. RaphaelFilgueiras
    • 2 years ago
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    decreasing f'<0 increasing f'>0

  20. VIbarguen1
    • 2 years ago
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    thank you so much

  21. RaphaelFilgueiras
    • 2 years ago
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    you are welcome

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