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Christos Group Title

Graphs and functions/derivatives, Hey can you help me with this? http://screencast.com/t/MVj3e9Ok2Yg0

  • one year ago
  • one year ago

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  1. .Sam. Group Title
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    \[p(x)=(x+1)^2(2x-x^2)\] From there, you can tell that (x+1)=0, x=-1 is a zero of p(x) Then factor this \((2x-x^2)\) and set the brackets equals zero and find the other 2 zeros

    • one year ago
  2. Christos Group Title
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    its -1 0 and 2

    • one year ago
  3. Christos Group Title
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    Oh so thats the first step

    • one year ago
  4. .Sam. Group Title
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    Yup |dw:1369655367345:dw|

    • one year ago
  5. .Sam. Group Title
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    Now, differentiate p(x), set it equals to zero to find the maximum and minimum points

    • one year ago
  6. Christos Group Title
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    its -4x^3+6x+2 if I am not mistaken we take x=0 and x=+-sqrt(3)/sqrt(2)

    • one year ago
  7. .Sam. Group Title
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    Yeah, set -4x^3+6x+2 equals zero and solve for x

    • one year ago
  8. Christos Group Title
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    yea I solved for x and I got 3 roots x=0 and x=+-sqrt(3)/sqrt(2) are my calculations correct??

    • one year ago
  9. .Sam. Group Title
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    Hmm, \[-4x^3+6x+2 \\ \\ -2(2x^3-3x-1) \\ \\ -2(x+1)(2x^2-2x-1)\] Factoring \(2x^2-2x-1\) you get \[x=\frac{1}{2} \left(1-\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\] These x are the max and min points of the graph, then use these x substitute into \[y=(x+1)^2(2x-x^2)\] Find the y's

    • one year ago
  10. Christos Group Title
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    hmm how did you get from line 2 to line 3?

    • one year ago
  11. Christos Group Title
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    http://screencast.com/t/yNU8TQxrMkn this here

    • one year ago
  12. .Sam. Group Title
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    I used trial an error with calculator, when \[p(-1)=-2(2(-1)^3-3(-1)-1)=0\] Then (x+1) is a factor of p(x), then use (x+1) divide \((2x^3-3x-1)\)

    • one year ago
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  13. Christos Group Title
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    oh ok lets find the ys

    • one year ago
  14. Christos Group Title
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    y = 0 ?

    • one year ago
  15. Christos Group Title
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    I subtitute with 0 or the roots?

    • one year ago
  16. .Sam. Group Title
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    Substitute these \(x=\frac{1}{2} \left(1-\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\) into \(y=(x+1)^2(2x-x^2)\)

    • one year ago
  17. .Sam. Group Title
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    Then that's the coordinates of max and minimum points

    • one year ago
  18. Christos Group Title
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    that was really long to find wasnt it? I tried to find the first one and I found (-2 + sqrt(3))/2 which seems to be wrong here

    • one year ago
  19. .Sam. Group Title
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    Yeah It's abit long For \(x=\frac{1}{2} \left(1-\sqrt{3}\right)\) or -0.3660, \[y=(x+1)^2(2x-x^2)\] \[y=(\frac{1-\sqrt3}{2} +1)^2(2(\frac{1-\sqrt3}{2})-(\frac{1-\sqrt3}{2})^2)\] \[y=\frac{9-6\sqrt{3}}{4}~~or-0.348\]

    • one year ago
  20. Christos Group Title
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    ok I see

    • one year ago
  21. .Sam. Group Title
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    (-0.3660,-0.3481) Then for the other coordinate is (1.366,4.848)

    • one year ago
  22. Christos Group Title
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    ok

    • one year ago
  23. Christos Group Title
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    Now that we know those coordinates how can we use them?

    • one year ago
  24. .Sam. Group Title
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    Yup |dw:1369657703814:dw|

    • one year ago
  25. .Sam. Group Title
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    Then the coefficient of highest degree is negative so the graph curves down|dw:1369657774047:dw|

    • one year ago
  26. .Sam. Group Title
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    And yeah, zeros are -1 0 and 2, why am I plotting 1 for.

    • one year ago
  27. Christos Group Title
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    hmm are you plotting it to get an estimation of the infliction point?

    • one year ago
  28. .Sam. Group Title
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    Yup

    • one year ago
  29. .Sam. Group Title
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    or else you won't know where it changes it's direction

    • one year ago
  30. Christos Group Title
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    I see , is there anything else left for this graph/exersise?

    • one year ago
  31. .Sam. Group Title
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    That's it

    • one year ago
  32. Christos Group Title
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    Thanks :) I will try to solve some on my own now, if I have any problem can I tell you?

    • one year ago
  33. .Sam. Group Title
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    I'll try :)

    • one year ago
  34. Christos Group Title
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    ok

    • one year ago
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