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lilfayfay

Let f be the function defined by f(x)=3x^5 - 5x^3 +2 (a) On what intervals is f increasing? (b) On what intervals is the graph of f concave upward? (c) Write the equation of each horizontal tangent line to the graph of f.

  • one year ago
  • one year ago

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  1. Goten77
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    1st derivative test?

    • one year ago
  2. lilfayfay
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    do first derivative test?

    • one year ago
  3. lilfayfay
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    @Goten77 ok for first derivative i got 15x^4 - 15x^2

    • one year ago
  4. Goten77
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    k thats right now factor it

    • one year ago
  5. Goten77
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    |dw:1358134355078:dw|

    • one year ago
  6. Goten77
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    set each = to 0 and get critical points |dw:1358134445508:dw| but also find where x never exist.. but it always does so ignore that part

    • one year ago
  7. Goten77
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    now set up the critical points to test for when its positive and negative and you use the 1st derivative when plugging in theese values|dw:1358134521331:dw| u just use values in between the numbers

    • one year ago
  8. Goten77
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    so thats part a when its increasing/decreasing.... also part c is just the graphs of y = 0 , -1, +1 since those are the 3 horizontal tangents

    • one year ago
  9. Goten77
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    part b is 2nd derivative test

    • one year ago
  10. Goten77
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    |dw:1358134818974:dw|

    • one year ago
  11. lilfayfay
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    for part a.) is it (-infinity,-1) & (1,infinity)

    • one year ago
  12. Goten77
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    yes

    • one year ago
  13. lilfayfay
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    ok for part b.) i got x=0, and \[\pm \frac{ \sqrt{2} }{ 2 }\]

    • one year ago
  14. Goten77
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    |dw:1358135080380:dw|

    • one year ago
  15. Goten77
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    |dw:1358135162509:dw| hmm you number works to...

    • one year ago
  16. lilfayfay
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    then i do the SDT? (Second Derivative Test)?

    • one year ago
  17. Goten77
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    well there the same i meant XD

    • one year ago
  18. Goten77
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    yes now you test those numbers between/around them

    • one year ago
  19. lilfayfay
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    ok, how do i do part c.)? o:

    • one year ago
  20. Goten77
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    i told u part c

    • one year ago
  21. Goten77
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    so um heres part b|dw:1358135773604:dw|

    • one year ago
  22. Goten77
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    hmm the squareroot number is very small....

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  23. lilfayfay
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    for b.) did u get \[(\frac{ \sqrt{2} }{ 2 }, \infty)\]

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  24. Goten77
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    oh i keep on thinking about point of inflection <.<

    • one year ago
  25. lilfayfay
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    |dw:1358128938034:dw| o.O

    • one year ago
  26. lilfayfay
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    is that drawing seem right? :\

    • one year ago
  27. Goten77
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    |dw:1358136164439:dw|

    • one year ago
  28. Goten77
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    yes your right... its only concave up on that last interval

    • one year ago
  29. lilfayfay
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    :"D

    • one year ago
  30. lilfayfay
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    so how do i do part c.) again sorry..

    • one year ago
  31. lilfayfay
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    wait so my part b.)was wrong?

    • one year ago
  32. Goten77
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    so heres what the graph looks like |dw:1358136607164:dw|

    • one year ago
  33. Goten77
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    aah i meant to say for part (c)

    • one year ago
  34. Goten77
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    k everything is right now.... i just got part letters mixed up

    • one year ago
  35. lilfayfay
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    so how do ido part c.)

    • one year ago
  36. Goten77
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    ......u kno what i meant... that entre paragraph just now is part C

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  37. Goten77
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    oh i said it kinda wrong earlier i meant part Cis when f(x)= -1,0,1 meaning plug those 3 values into the original function and thats when the graph has a tangent lind so then you get y= 4 y= 2 y= 0 and thats when the graph has the horizontal tangents the x values of -1,0,1 came from the 1st derivative test since those are the critical points meaning where the derivative = 0

    • one year ago
  38. lilfayfay
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    ah ok ty so for part c.) i got y=-2x+2

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  39. Goten77
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    no i gave u part 2

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  40. Goten77
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    part c*

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  41. lilfayfay
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    yeah it says write the equation

    • one year ago
  42. Goten77
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    ill explain it again

    • one year ago
  43. Goten77
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    k so the 1st derivative gives us critical points meaning where the slope = 0 also meaning at those points on the original graph is where it has a horizontal tangent..... so taht means the horizontal tangents are in the form of y = # and thats it so our critical points were -1 0 and 1 plug those into the original equasion and get y= 4 y= 2 y= 0 and thats it .. those are your 3 horizontal tangents

    • one year ago
  44. lilfayfay
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    ohso my answer waswrong x-x

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  45. lilfayfay
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    okty :P

    • one year ago
  46. lilfayfay
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    @Goten77 canu help me with another 2 more pls? ^^

    • one year ago
  47. Goten77
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    sure

    • one year ago
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