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anonymous

  • one year ago

Help please?

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  1. anonymous
    • one year ago
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    3 turning points, should be interesting!

  2. anonymous
    • one year ago
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    I have no idea how to start it. Should I derive it first and then find the values?

  3. anonymous
    • one year ago
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    I got 0 and -3

  4. anonymous
    • one year ago
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    @mukushla was right, f'(x)=0 at the turning point

  5. anonymous
    • one year ago
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    @DanJS

  6. DanJS
    • one year ago
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    can you find \[\frac{ d }{ dx }f(x)\]

  7. DanJS
    • one year ago
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    that expression will give you the slope of any tangent line to the curve at a point

  8. anonymous
    • one year ago
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    When I derive it I got 6x^2+12x

  9. DanJS
    • one year ago
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    You just have to remember, \[\frac{ d }{ dx }[x^n]= n*x ^{n-1}\]

  10. DanJS
    • one year ago
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    right, now set that slope expression to what they want, zero

  11. DanJS
    • one year ago
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    horizontal line, zero slope

  12. anonymous
    • one year ago
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    Dan the man!

  13. DanJS
    • one year ago
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    (6x)(x+2) = 0 x can be 2 values , each quantity zero

  14. anonymous
    • one year ago
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    Factors right?

  15. DanJS
    • one year ago
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    6x=0 or x+2=0 horizontal tangent line

  16. DanJS
    • one year ago
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    use those x values in the original f(x) and find out each corresponding y value (x,y)

  17. DanJS
    • one year ago
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    or nvermind, they dont want it

  18. anonymous
    • one year ago
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    You get all 3 values Dan?

  19. DanJS
    • one year ago
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    it is a cubic function , it turns around 2 times,

  20. anonymous
    • one year ago
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    So it would be -2 and 0?

  21. DanJS
    • one year ago
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    http://awesomescreenshot.com/0b9514fn7e

  22. DanJS
    • one year ago
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    right

  23. DanJS
    • one year ago
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    Take first derivative, set that expression to zero, solve for the values of x for each tangent line if they exist

  24. anonymous
    • one year ago
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    Alright thank you so much!

  25. DanJS
    • one year ago
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    np

  26. anonymous
    • one year ago
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    as usual, well done @DanJS

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spraguer (Moderator)
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