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Christos

  • 2 years ago

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

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  1. .Sam.
    • 2 years ago
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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

  2. Christos
    • 2 years ago
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    its -1 0 and 2

  3. Christos
    • 2 years ago
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    Oh so thats the first step

  4. .Sam.
    • 2 years ago
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    Yup |dw:1369655367345:dw|

  5. .Sam.
    • 2 years ago
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    Now, differentiate p(x), set it equals to zero to find the maximum and minimum points

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

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

  8. Christos
    • 2 years ago
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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??

  9. .Sam.
    • 2 years ago
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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

  10. Christos
    • 2 years ago
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    hmm how did you get from line 2 to line 3?

  11. Christos
    • 2 years ago
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    http://screencast.com/t/yNU8TQxrMkn this here

  12. .Sam.
    • 2 years ago
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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 \((2x^3-3x-1)\)

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  13. Christos
    • 2 years ago
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    oh ok lets find the ys

  14. Christos
    • 2 years ago
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    y = 0 ?

  15. Christos
    • 2 years ago
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    I subtitute with 0 or the roots?

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

  17. .Sam.
    • 2 years ago
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    Then that's the coordinates of max and minimum points

  18. Christos
    • 2 years ago
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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

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

  20. Christos
    • 2 years ago
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    ok I see

  21. .Sam.
    • 2 years ago
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    (-0.3660,-0.3481) Then for the other coordinate is (1.366,4.848)

  22. Christos
    • 2 years ago
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    ok

  23. Christos
    • 2 years ago
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    Now that we know those coordinates how can we use them?

  24. .Sam.
    • 2 years ago
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    Yup |dw:1369657703814:dw|

  25. .Sam.
    • 2 years ago
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    Then the coefficient of highest degree is negative so the graph curves down|dw:1369657774047:dw|

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

  27. Christos
    • 2 years ago
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    hmm are you plotting it to get an estimation of the infliction point?

  28. .Sam.
    • 2 years ago
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    Yup

  29. .Sam.
    • 2 years ago
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    or else you won't know where it changes it's direction

  30. Christos
    • 2 years ago
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    I see , is there anything else left for this graph/exersise?

  31. .Sam.
    • 2 years ago
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    That's it

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

  33. .Sam.
    • 2 years ago
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    I'll try :)

  34. Christos
    • 2 years ago
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    ok

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