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anonymous

  • one year ago

How do you simplify this equation into a value for y ? y = (Sqrt[2]/2) / E^((Sqrt[2]/2)^2) ​

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  1. UsukiDoll
    • one year ago
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    oh wow what is this lol

  2. anonymous
    • one year ago
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    y = (Sqrt[2]/2) / E^((Sqrt[2]/2)^2) I think it's this.. ​\[y =\frac{ \frac{\sqrt{2}}{2} }{ E^{\frac{\sqrt{2}}{2}^2} ​}\]

  3. anonymous
    • one year ago
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    Im supposed to be able to do this by hand... :/

  4. UsukiDoll
    • one year ago
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    OMG who assigns this?

  5. anonymous
    • one year ago
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    its the worst calculus course ever invented.. bill davis, horacio porta and Jerry Uhl

  6. anonymous
    • one year ago
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    apparently a flipped curriculum.. where they explain nothing, and try to have the students work it out for themselves.

  7. UsukiDoll
    • one year ago
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    ffffffffffffffffffffffff

  8. UsukiDoll
    • one year ago
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    \[(\frac{\sqrt{2}}{2})^2\] I would start expanding this first.

  9. UsukiDoll
    • one year ago
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    how is this Calculus? More like torture.

  10. anonymous
    • one year ago
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    maybe \[ (\frac{2^.5} { 2 }) (\frac{2^.5}{ 2 })\] \[ \frac{(2^.5)(2^.5)} { 4 } \]

  11. anonymous
    • one year ago
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    Ive been thinking the same thing.. lol

  12. UsukiDoll
    • one year ago
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    ummm ... \[\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2}\]

  13. anonymous
    • one year ago
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    woah.. is that true?

  14. anonymous
    • one year ago
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    yes it is

  15. UsukiDoll
    • one year ago
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    \[\LARGE e^{\frac{1}{2}}\]

  16. UsukiDoll
    • one year ago
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    darn it ... I sort of forgot some of the e rules I know for \[\large e^{2+x} \rightarrow e^2e^x\]

  17. anonymous
    • one year ago
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    so I think I can move the 2 in the numerator down to the denominator

  18. UsukiDoll
    • one year ago
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    I need to relook it up... I think it involves one of em .

  19. anonymous
    • one year ago
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    yes thats true..

  20. UsukiDoll
    • one year ago
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    can't remember =/

  21. myininaya
    • one year ago
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    hey so you are suppose to "simplify" the expression called y? and y is given as: \[y=\frac{\frac{\sqrt{2}}{2}}{e^{(\frac{\sqrt{2}}{2})^2}}\]

  22. UsukiDoll
    • one year ago
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    yeah and I already expanded the bottom

  23. UsukiDoll
    • one year ago
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    because that's an easy place to start

  24. anonymous
    • one year ago
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    so I think we get y= 2^(1/2) / 2 E^(1/2)

  25. UsukiDoll
    • one year ago
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    where did that 2 come from? ^ at the denominator

  26. anonymous
    • one year ago
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    btw is there an easy way to enter these equations? that equation tool takes me forever.

  27. myininaya
    • one year ago
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    you can also write that as: \[y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}}\]

  28. myininaya
    • one year ago
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    you can also chose to multiply the top and bottom by e^(1/2) to rationalize the denominator

  29. myininaya
    • one year ago
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    well sorta rationalize anyways since e isn't rational :p

  30. UsukiDoll
    • one year ago
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    \[\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}} \] I rather have this XD can't we flip the second fraction over or use one of the exponent laws for that 1/e^{something}

  31. anonymous
    • one year ago
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    I cant believe these questions aren't all over open study.. I must be the only idiot in the world doing this course..

  32. UsukiDoll
    • one year ago
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    cough common core cough xD

  33. myininaya
    • one year ago
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    yes you can do this: \[y=\frac{\sqrt{2}}{2 e^\frac{1}{2}} \cdot \frac{e^\frac{1}{2}}{e^\frac{1}{2}}\]

  34. UsukiDoll
    • one year ago
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    wasn't there also an ... I forgot what's it's called but if I have something like this... I can split it up...what's it called? \[\large e^{2+x} \rightarrow e^2e^x \]

  35. myininaya
    • one year ago
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    that is one the law of exponents

  36. anonymous
    • one year ago
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    I have this in notes... The main laws of exponents are e^(a + b) = e^a e^b e^(a - b) = e^a/e^b (e^a)^b = e^(a b) . a / E^n = a E^-n negative exponent law

  37. UsukiDoll
    • one year ago
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    ah... so would this work? \[\large e^{\frac{1}{2}} \rightarrow e^{1}e^{-2}\]

  38. myininaya
    • one year ago
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    no

  39. myininaya
    • one year ago
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    \[e^1e^{-2}=e^{1-2}=e^{-1}=\frac{1}{e}\]

  40. UsukiDoll
    • one year ago
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    oh xD now I remember

  41. myininaya
    • one year ago
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    have you considered getting rid of the square root part on bottom as I suggested earlier ?

  42. UsukiDoll
    • one year ago
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    e and the other exponent laws are different \[n^{a-b} = \frac{n^a}{n^b}\]

  43. myininaya
    • one year ago
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    n can be e

  44. UsukiDoll
    • one year ago
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    yeah \[\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2} \]

  45. UsukiDoll
    • one year ago
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    oh whoops yeah for something else not the exponent law thing

  46. anonymous
    • one year ago
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    maybe log.. can be applied?

  47. myininaya
    • one year ago
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    remember this: \[y=\frac{\sqrt{2}}{2 e^\frac{1}{2}} \cdot \frac{e^\frac{1}{2}}{e^\frac{1}{2}} \\ y= \frac{\sqrt{2} e^\frac{1}{2}}{2 e}=\frac{\sqrt{2 e}}{2e}\]

  48. myininaya
    • one year ago
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    anyways that is the form I would leave it in but you may think the square thing looks better on bottom and if you think that you can multiply top and bottom by sqrt(2e) and simplify from there but this form above is the one I would leave it as

  49. myininaya
    • one year ago
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    square root thing*

  50. anonymous
    • one year ago
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    My calculator says this will end up being 1/Sqrt[2E] and I would get a value of 0.428882

  51. myininaya
    • one year ago
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    ok if you want that other result as I said you can put the square root thing in the bottom by multiplying top and bottom by sqrt(2e) and simplifying from there

  52. anonymous
    • one year ago
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    are we on track?

  53. myininaya
    • one year ago
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    \[y=\frac{\sqrt{2e}}{2e} \\ \text{ this is my preferred form } \text{ multiply } \sqrt{2e} \text{ on bottom and \top and simplify from } \\ \text{ there if you want } y=\frac{1}{\sqrt{2e}}\]

  54. dan815
    • one year ago
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    sounds like a fun course :)

  55. anonymous
    • one year ago
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    lol, dan you would own this course.. :)

  56. anonymous
    • one year ago
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    is there a trick to turning this into a value that I can set into a chart.. I'm supposed to plot points with this, by hand.

  57. myininaya
    • one year ago
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    example: \[\frac{\sqrt{u}}{u} \\ \frac{\sqrt{u}}{u} \cdot \frac{\sqrt{u}}{\sqrt{u}} =\frac{u}{u \sqrt{u}}=\frac{1}{\sqrt{u}}\]

  58. anonymous
    • one year ago
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    ah I like that perspective.. thnx. myininaya, I will add that to the list..

  59. UsukiDoll
    • one year ago
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    oh I see.. after we got \[e^\frac{1}{2}\] we changed it in radical form and rationalize!

  60. dan815
    • one year ago
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    |dw:1434788368933:dw|

  61. myininaya
    • one year ago
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    @hughfuve I still don't understand why you wanted to write it in this particular form

  62. UsukiDoll
    • one year ago
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    \[\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{e}} \] \[\frac{\sqrt{2}}{2\sqrt{e}}\] \[\frac{\sqrt{2}}{2\sqrt{e}} \cdot \frac{\sqrt{e}}{\sqrt{e}}\] \[\frac{\sqrt{2e}}{2e}\]

  63. dan815
    • one year ago
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    u wanna graph it? it would be a horizontal line

  64. myininaya
    • one year ago
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    And is there a bigger question? You say you want to graph this? this is y=constant where we know the constant is between 0 and 1 since we had \[\frac{\sqrt{2e}}{2e} \text{ where } 0<\sqrt{2e}<2e \text{ dividing both sides by } 2e \text{ gives } \\ 0<\frac{\sqrt{2e}}{2e}<1\]

  65. anonymous
    • one year ago
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    Oh Im just trying to get it to a place, where I can calculate a number , I need to arrive at 0.428882

  66. anonymous
    • one year ago
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    and do it by hand... there are no calculators allowed.

  67. UsukiDoll
    • one year ago
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    O_O!

  68. dan815
    • one year ago
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    u can use those series to approximate e

  69. anonymous
    • one year ago
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    well, that's just a single point.. on a graph, but it's a critical point.. it's the maxima

  70. dan815
    • one year ago
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    writing out the answer as y=1/sqrt(2e), is the nicest way imo

  71. anonymous
    • one year ago
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    this equation is the 1st derivative[0]

  72. myininaya
    • one year ago
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    Can you give the whole question?

  73. anonymous
    • one year ago
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    okay... 1 sec.. typing it in..

  74. anonymous
    • one year ago
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    Examine the derivative of f[x] = x E^(-x^2) to give a reasonably good hand sketch of the curve y = f[x] on the axes below. What are the maximum and minimum values f[x] can have ?

  75. UsukiDoll
    • one year ago
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    first derivative will have something like y' = ....

  76. UsukiDoll
    • one year ago
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    oh this is another question...

  77. anonymous
    • one year ago
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    this is where this equation came from... I worked out the derivative.. and then put it in as x for f[x]

  78. myininaya
    • one year ago
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    so when you set f'=0 you got 1-2x^2=0 right?

  79. anonymous
    • one year ago
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    maybe I was completely on the wrong track though.

  80. anonymous
    • one year ago
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    argh.. I ballsed up there hey?

  81. anonymous
    • one year ago
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    I had this...

  82. anonymous
    • one year ago
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    f'[x] = -2x^2 E^(-x^2) + E^(-x^2)

  83. myininaya
    • one year ago
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    right and you can factor the exp(-x^2) out

  84. anonymous
    • one year ago
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    0 = E^(-x^2) -2x^2 E^(-x^2)

  85. anonymous
    • one year ago
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    head slap!

  86. myininaya
    • one year ago
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    \[f'=e^{-x^2}(1-2x^2) \\ e^{-x^2} \neq 0 \text{ so we only have \to solve } 1-2x^2=0\]

  87. anonymous
    • one year ago
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    I see...

  88. anonymous
    • one year ago
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    looking at the equation now and it's jumping out at me.

  89. myininaya
    • one year ago
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    but looking at \[f(x)=xe^{-x^2} \ \\ f(\frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}e^{-(\frac{1}{\sqrt{2}})^2}=\frac{1}{\sqrt{2}}e^{-\frac{1}{2}}=\frac{1}{\sqrt{2} e^\frac{1}{2}}=\frac{1}{\sqrt{2e}} \\ \text{ you already got this \above } \\ \text{ now you can find the other value } f(\frac{-1}{\sqrt{2}})\]

  90. myininaya
    • one year ago
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    but you still need to determine if they are max,min,neither

  91. myininaya
    • one year ago
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    e^(-x^2) is positive for all x (1-2x^2) is a parabola; it is what will tell us if the derivative is positive or negative

  92. anonymous
    • one year ago
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    2nd derivative? or is there a better way

  93. myininaya
    • one year ago
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    this is a rough graph of y=1-2x^2 |dw:1434789467495:dw| we are just using this graph to determine when f' is positive or negative you know where f' is: \[f'(x)=e^{-x^2}(1-2x^2)\]

  94. UsukiDoll
    • one year ago
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    2nd derivative is concavity.

  95. myininaya
    • one year ago
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    looking at this graph can you tell me for what intervals y is positive and what intervals y is negative

  96. anonymous
    • one year ago
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    somewhere between +/- 1/2 and 1/1

  97. anonymous
    • one year ago
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    .5 .. .7 ... 1 ?

  98. anonymous
    • one year ago
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    sqrt 2 = about 1.4 ?

  99. UsukiDoll
    • one year ago
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    1/1 = 1 though

  100. myininaya
    • one year ago
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    I'm trouble understanding what you are saying ... umm... in the picture I drew of the parabola y=1-2x^2 y is positive when the graph is above the x-axis it is positive on (-1/sqrt(2),1/sqrt(2)) y is negative when the graph is below the x-axis it is negative on (-inf,-1/sqrt(2)) U (1/sqrt(2),inf)

  101. anonymous
    • one year ago
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    thats why I was thinking ... 1/1 ... 1/sqrt(2) ... 1/2 somewhere in the middle.

  102. anonymous
    • one year ago
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    and came up with about .7

  103. UsukiDoll
    • one year ago
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    sqrt(2) = 1.41 1... 1/1.41 = 0.70 1/2 = 0.5

  104. anonymous
    • one year ago
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    or do we only need to know if positive or negative?

  105. anonymous
    • one year ago
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    on the 2nd derivative.. I just learned that if f[x] is a possible extrema and f''[x] =0 then it is not maxima or minima, but if f''[x] = negative then f[x] is a maxima, and if f''[x] is positive then f[x] is a minima.. I think its a Fermat law ...

  106. anonymous
    • one year ago
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    perl suggested it earlier on a problem.

  107. myininaya
    • one year ago
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    this means: \[f'(x)=e^{-x^2}(1-2x^2) \\ f' \text{ is positive on } (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} ) \text{ which means } f \text{ is increasing there } \\ f' \text{ is negative on } (-\infty, \frac{-1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \text{ which means } f \\ \text{ is decreasing there }\]

  108. anonymous
    • one year ago
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    so we have our parabola orientation. sweet.

  109. myininaya
    • one year ago
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    anyways so lets make a little sign chart here for the derivative thing: if it switches from increasing to decreasing at a critical number then at that critical number we have a max if it switches from decreasing to increasing at a critical number then at that critical number we have a min |dw:1434790458729:dw|

  110. myininaya
    • one year ago
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    so using this can you determine which of f(-1/sqrt(2)) or f(1/sqrt(2)) will give us a max and which will give us a min?

  111. myininaya
    • one year ago
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    remember the f'=- means f decreasing and f'=+ means f increasing

  112. anonymous
    • one year ago
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    yes min in the -x max in the +x

  113. myininaya
    • one year ago
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    right so the min value is f(-1/sqrt(2)) and the max value is f(1/sqrt(2))

  114. anonymous
    • one year ago
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    / \ 0 \ /

  115. myininaya
    • one year ago
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    so we know our graph looks something like this between -1/sqrt(2) and 1/sqrt(2) |dw:1434790667090:dw|

  116. myininaya
    • one year ago
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    now the other thing we know our function is decreasing on the other intervals ...so we know we have something like this for outside that interval: |dw:1434790765867:dw| I didn't finish the tail ends because we don't know the end behavior of the graph yet

  117. anonymous
    • one year ago
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    and those points will be at.. y = f[ +/- 1/Sqrt[2] ]

  118. myininaya
    • one year ago
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    I ask you to consider the following limits for end behavior : \[\lim_{x \rightarrow \infty}f(x) \text{ and } \lim_{x \rightarrow -\infty}f(x)\]

  119. anonymous
    • one year ago
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    oh, actually..they gave me the points for that.. so we can skip that..

  120. anonymous
    • one year ago
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    its on a plot... two points near + / - (2,0)

  121. myininaya
    • one year ago
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    and I didn't mean to draw a 1 at those points by the way let me redo that

  122. myininaya
    • one year ago
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    |dw:1434790888537:dw|

  123. myininaya
    • one year ago
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    |dw:1434790920439:dw| you still need to finish the tail ends of the graphs where are they decreasing from do you think

  124. myininaya
    • one year ago
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    decreasing from and to *

  125. anonymous
    • one year ago
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    nice

  126. anonymous
    • one year ago
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    if you wanted now to work out the square root of 2E by hand.. is there a trick? other than trial and error and squeezing ?

  127. myininaya
    • one year ago
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    you can use a calculator ... or...you can say this: \[2<e<3 \\ 4 <2e<6 \\ \sqrt{4}< \sqrt{2e}<\sqrt{6} \approx 2.5 \\ 2<\sqrt{2e}<2.5\] but earlier on don't anything wrong with the way I approximated our sqrt(2e)/(2e) I just said it is between 0 and 1 since we know: \[0<\sqrt{2e}<2e \\ \ \text{ now dividing by } 2e \text{ we have } 0<\frac{\sqrt{2e}}{2e}<1\] just place it somewhere between 0 and 1 as I did in the picture above

  128. anonymous
    • one year ago
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    nice

  129. myininaya
    • one year ago
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    and remember sqrt(2e)/(2e) is the same number as 1/sqrt(2e) we just wrote it differently

  130. anonymous
    • one year ago
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    gotcha

  131. myininaya
    • one year ago
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    summary: \[f(x)=xe^{-x^2} \\ f'(x)=e^{-x^2}+x(-2x)e^{-x^2} \\ f'(x)=e^{-x^2}(1-2x^2) \\ \text{ we know } e^{-x^2} >0 \text{ for all } x \\ 1-2x^2=0 \text{ when } x=\pm \frac{1}{\sqrt{2}} \text{ or if you prefer \to write as } x=\pm \frac{\sqrt{2}}{2} \\ \text now f(\pm \frac{\sqrt{2}}{2})=\pm \frac{1}{\sqrt{2e}} \text{ or if you prefer } \pm \frac{\sqrt{2e}}{2e} \\ \text{ also looking at } f' \\ \text{ we have } f'>0 \text{ when } x \in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}) \text{ so } f \text{ is increasing on } (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) \\ \text{ we have } f'<0 \text{ when } x \in (-\infty,\frac{-1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \\ \text{ so } f \text{ is decreasing on } (-\infty,\frac{-1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \\ \text{ so this tells us our max } f(\frac{\sqrt{2}}{2})=\frac{1}{\sqrt{2e}} \text{ and our min is } f(-\frac{\sqrt{2}}{2})=-\frac{1}{\sqrt{2e}} \] so from this information we can roughly sketch part of the graph: |dw:1434791775024:dw| the only left for you is to work out the end behavior the graph

  132. anonymous
    • one year ago
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    oh thats amazing help.. you out did yourself here.. I got those end behaviours.. this equation is decaying to 0 because of that exponential denominator f[x] = x/ E^(x^2)

  133. myininaya
    • one year ago
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    you probably could use the fact your function is odd to justify the end behavior of your graph as: as we approach pos/neg infinity the graph is getting closer to y=0

  134. myininaya
    • one year ago
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    |dw:1434792046243:dw| yep so you would end up with something like this

  135. anonymous
    • one year ago
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    gotcha.. phew! exactly what I was thinking..

  136. anonymous
    • one year ago
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    man, I wish every problem wasn't like this. I got 11 more on this assignment.. just like this one.. lol.

  137. anonymous
    • one year ago
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    but this will get me a long way to solving the rest..

  138. myininaya
    • one year ago
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    good luck on the rest

  139. myininaya
    • one year ago
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    myininaya is probably going to have to go now unless you have anymore questions about this particular question?

  140. anonymous
    • one year ago
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    thank you thank you again.. I hope the rest of your night are fun problems.

  141. anonymous
    • one year ago
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    Im good

  142. myininaya
    • one year ago
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    I actually wasted my whole. I should have been in bed like hours ago. It is 4:25 AM here.

  143. myininaya
    • one year ago
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    my whole night*

  144. anonymous
    • one year ago
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    lol.. oh wow.. well I appreciate your time very very much, you're a life saver.

  145. myininaya
    • one year ago
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    I wouldn't say wasted. I had fun. Played some video games and did some math problems. Fun fun.

  146. myininaya
    • one year ago
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    anyways goodnight(sort of )

  147. anonymous
    • one year ago
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    here's a sneak peak at my next challenge.. please dont answer ... To Plot this by hand ... f[x] = (x^2 - 2 Log[x])/2

  148. anonymous
    • one year ago
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    sigh.. going to be a long night. lol.

  149. myininaya
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtI.aspx This site maybe helpful too. Not sure if you ever heard of it.

  150. UsukiDoll
    • one year ago
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    at least @hughfuve got his moneys worth and Qualified Help unlike the last time I came on one of his QH questions.

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