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anonymous

  • 5 years ago

Find the local extrema and inflection points of g(x)=e^(-x^2/2).

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  1. amistre64
    • 5 years ago
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    is that (-x^2)/2 or is it -x^(2/2) ? its probably the first, but the first rule of mathematics is NEVER trust a mathmetician :)

  2. amistre64
    • 5 years ago
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    dy/dx = dy/du * du/dx y(u(x)) = e^((-x^2)/2) where y=e^u and u=-x^2/2.

  3. amistre64
    • 5 years ago
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    dy/du = e^u du/dx = -2x/2 = -x y' = -x * e^((-x^2)/2) thats the first derivative.

  4. amistre64
    • 5 years ago
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    y'' the second derivative tells us y' is max or min; and also were the inflection point are: y'' = (-) MAX y'' = (+) MIN y'' = 0 inflection

  5. anonymous
    • 5 years ago
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    it is (-x^2)/2~

  6. amistre64
    • 5 years ago
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    y'' you will need to find by the product rule: fl' + f'l

  7. amistre64
    • 5 years ago
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    y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]

  8. anonymous
    • 5 years ago
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    for g'(x)=-xe^(-x^2/2), how to i find x?

  9. amistre64
    • 5 years ago
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    make g'(x) equal to 0 to find your critical points...the bends of the graph.

  10. anonymous
    • 5 years ago
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    i just don;t know how to find the value of x in e^(-x^2/2) :(

  11. anonymous
    • 5 years ago
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    when i equated the equation to zero

  12. amistre64
    • 5 years ago
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    x=0 I am thinking there is no value for e^x that will make it 0. Think about the graph of the exponent funtion, it never reaches 0

  13. amistre64
    • 5 years ago
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    do you agree?

  14. anonymous
    • 5 years ago
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    yes, coz i cant think of any value that can make e to zero

  15. amistre64
    • 5 years ago
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    since e^x cannot ever reach 0 there is no value of "x" that will make it = 0....so it doesnt matter :)

  16. anonymous
    • 5 years ago
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    so how would the graph look like? would it be like concave up?

  17. amistre64
    • 5 years ago
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    so we have a point at 0 that tells us there is something going on at x=0. lets use the second derivative to get more information

  18. anonymous
    • 5 years ago
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    concave up from (-INF,-1), concave down (-1,1), concave up (1,INF)?

  19. anonymous
    • 5 years ago
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    and the local max. is at x=0?

  20. amistre64
    • 5 years ago
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    There is a bend at 0.... but we dont really know what it is doing until we use the g''(x) to tell us more about what is happening at x=0 y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)] we can clean this up some: y'' = x^2 * e^((-x^2)/2) - e^((-x^2)/2)

  21. amistre64
    • 5 years ago
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    when x = 0; y'' = -e^(...) y'' is (-) at x=0 which tells us that it is a MAX

  22. amistre64
    • 5 years ago
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    when is y'' = 0? for our inflections. At x=1, cause e^(...) - e(..) = 0 so, at x=1 we have a change of direction.

  23. amistre64
    • 5 years ago
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    so, heres what we know: at g(0) there is a Max; and at g(1) there is an inflection. plug in those value for x in the original to get your y spots. Does this make sense?

  24. anonymous
    • 5 years ago
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    ah~yes! thanks!

  25. anonymous
    • 5 years ago
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    f(x)=x^3+y^3-3xy=0 (this is not a function) so do i need to use implicit differentiation?

  26. anonymous
    • 5 years ago
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    or i just differentiate it normally? it will be like =f'(x)=3x^2-3x=0

  27. amistre64
    • 5 years ago
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    sure. is it "y" is a function of "x" implicitly? cause it could go the other way around; or even be implied by an external variable like time

  28. amistre64
    • 5 years ago
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    you show me how you think the steps go; and I will let you kow if you are right...

  29. anonymous
    • 5 years ago
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    f(x)=x^3+y^3-3xy=0 f'(x)=3x^2+0-3x

  30. amistre64
    • 5 years ago
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    explain to me: (d/dx)(y^3) = 0 cause thats an error.

  31. anonymous
    • 5 years ago
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    for this, i treated y as a constant and i didn't use implicit differentiation. should i use it? if i use it, f'(x)=3x^2+3y^2*y'-3y*y'*x

  32. amistre64
    • 5 years ago
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    yes it; but your last term is in error... look at it again.

  33. anonymous
    • 5 years ago
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    woops! wrong

  34. anonymous
    • 5 years ago
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    yes haha lemme correct it 3(y+y'x)

  35. amistre64
    • 5 years ago
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    YES!! - now solve this equation for y'

  36. amistre64
    • 5 years ago
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    f'(x) = y' So get all you y' to one side and everything else to the other side; factor out the y' and then get y' by itself basic algebra stuff

  37. anonymous
    • 5 years ago
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    y'=3x^2+y'(3y^2-3x)-3y y'-y'(3y^2-3x)=3x^2-3y y'(1-3y^2+3x)=3x^2-3y y'=(3x^2-3y)/(1-3y^2+3x)

  38. amistre64
    • 5 years ago
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    as long as you kept track of yourself; you got it :)

  39. anonymous
    • 5 years ago
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    so do i need to find the x values in terms of y?

  40. anonymous
    • 5 years ago
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    equate the whole equation to zero?

  41. amistre64
    • 5 years ago
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    nope; they will give you values for x and y; or values for y' and x; or values for y' and y. you just plug in what they give you and solve for the unknown that they dont.

  42. anonymous
    • 5 years ago
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    but they did nt give me any values..: sketch the curve x^3+y^3-3xy=0, indicating all extrema and points of inflection. Be sure to show all of your work.

  43. amistre64
    • 5 years ago
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    you do the same thing for "with respect to time" but the thing is (d/dt) doesnt vanish, so you keep your (dx/dt) and (dy/dt). ahh, you solve the original for y. that can be done explicitly becasue you can solve for y.

  44. amistre64
    • 5 years ago
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    or use what we got :)

  45. amistre64
    • 5 years ago
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    little hint; y' = 0 when that top part = 0...try that

  46. anonymous
    • 5 years ago
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    x=+/- sqrty?

  47. amistre64
    • 5 years ago
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    3x^2 -3y = 0 when?

  48. amistre64
    • 5 years ago
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    3x^2 = 3y y=x^2 when does n^2 = n?

  49. anonymous
    • 5 years ago
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    when n=-n or +n?

  50. anonymous
    • 5 years ago
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    sqrt

  51. anonymous
    • 5 years ago
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    so x=+/- sqrty?

  52. amistre64
    • 5 years ago
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    (0)^2 = 0 and (1)^2 = 1

  53. amistre64
    • 5 years ago
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    do these make the bottom go to 0? if so, toss whichever one does.

  54. amistre64
    • 5 years ago
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    but as is, we have bends at (0,0 and (1,1) does that make sense to you?

  55. anonymous
    • 5 years ago
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    yes!

  56. amistre64
    • 5 years ago
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    to find the y'' just derive each side again with respect to x

  57. amistre64
    • 5 years ago
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    your gonna have to use the quotient rule; and keep track of yourself tho :)

  58. anonymous
    • 5 years ago
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    thats such a longgggggggggggggggggg equation...

  59. amistre64
    • 5 years ago
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    lol..... do it, and eat your brussels sprouts :)

  60. amistre64
    • 5 years ago
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    top is: (1 -3y^2 +3x)(6x -3y')-(3x^2 -3y)(-6yy' +3)

  61. amistre64
    • 5 years ago
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    the bottom is: (1 -3y^2 +3x)(1 -3y^2 +3x)

  62. amistre64
    • 5 years ago
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    im going with you work...so if its wrong... :)

  63. anonymous
    • 5 years ago
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    y"= [-3y'(1+5y^2+3x+6x^2y)+3x(2-6y^2+9x)-9y]/(1-3y^2+3x^2)^2

  64. anonymous
    • 5 years ago
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    uhhh i guess its right coz my eyes even got dried when i was differentiating it lol

  65. amistre64
    • 5 years ago
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    we know y'=0 when we plug in(0,0) or (1,1) right? so plug all that in and see it we are (-),(+), or (0)

  66. amistre64
    • 5 years ago
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    it might be easier to work with the unfactored forms :)

  67. amistre64
    • 5 years ago
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    it is it is..... at (0,0) y'' = -3 we got a max at (0,0)

  68. amistre64
    • 5 years ago
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    at (1,1) we get y''=+6 we have a MIN at (1,1)

  69. anonymous
    • 5 years ago
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    i got y"=-3y' when (0,0)

  70. anonymous
    • 5 years ago
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    but if i plug in 0 for y', it will be zero!

  71. amistre64
    • 5 years ago
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    use the stuff I put up, its alot easier to work it in chunks that way

  72. amistre64
    • 5 years ago
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    less degree for error

  73. anonymous
    • 5 years ago
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    okay! but why it would be max when y"=-3? wouldnt it be min?

  74. anonymous
    • 5 years ago
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    comparing to y"=+6

  75. amistre64
    • 5 years ago
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    (1)(0) - (0)(3) ---------- (1)(1)

  76. amistre64
    • 5 years ago
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    we good at (0,0) = y'' = -3?

  77. anonymous
    • 5 years ago
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    did you plug in y'=0?

  78. amistre64
    • 5 years ago
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    thin of it like this: y = -(x^2) is an upside down bowl: y' = -(2x) y'' = -(2) = MAX

  79. amistre64
    • 5 years ago
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    yes, I plug in (0,0) and y'=0 into what "I" did :)

  80. anonymous
    • 5 years ago
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    but if i also plug in y'=0, it would be like: [(0)(1)+(0)(2)-0]/ (1)(1), so wouldnt it be y"=0?

  81. anonymous
    • 5 years ago
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    i undersand the max thing now btw:) thanks

  82. amistre64
    • 5 years ago
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    .... i hate to say this; but I trust my version of it better. You might have made an error when trying to squash everything together. In fact, I am sure of it :)

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