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

1. UsukiDoll

oh wow what is this lol

2. anonymous

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

Im supposed to be able to do this by hand... :/

4. UsukiDoll

OMG who assigns this?

5. anonymous

its the worst calculus course ever invented.. bill davis, horacio porta and Jerry Uhl

6. anonymous

apparently a flipped curriculum.. where they explain nothing, and try to have the students work it out for themselves.

7. UsukiDoll

ffffffffffffffffffffffff

8. UsukiDoll

$(\frac{\sqrt{2}}{2})^2$ I would start expanding this first.

9. UsukiDoll

how is this Calculus? More like torture.

10. anonymous

maybe $(\frac{2^.5} { 2 }) (\frac{2^.5}{ 2 })$ $\frac{(2^.5)(2^.5)} { 4 }$

11. anonymous

Ive been thinking the same thing.. lol

12. UsukiDoll

ummm ... $\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2}$

13. anonymous

woah.. is that true?

14. anonymous

yes it is

15. UsukiDoll

$\LARGE e^{\frac{1}{2}}$

16. UsukiDoll

darn it ... I sort of forgot some of the e rules I know for $\large e^{2+x} \rightarrow e^2e^x$

17. anonymous

so I think I can move the 2 in the numerator down to the denominator

18. UsukiDoll

I need to relook it up... I think it involves one of em .

19. anonymous

yes thats true..

20. UsukiDoll

can't remember =/

21. myininaya

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

yeah and I already expanded the bottom

23. UsukiDoll

because that's an easy place to start

24. anonymous

so I think we get y= 2^(1/2) / 2 E^(1/2)

25. UsukiDoll

where did that 2 come from? ^ at the denominator

26. anonymous

btw is there an easy way to enter these equations? that equation tool takes me forever.

27. myininaya

you can also write that as: $y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}}$

28. myininaya

you can also chose to multiply the top and bottom by e^(1/2) to rationalize the denominator

29. myininaya

well sorta rationalize anyways since e isn't rational :p

30. UsukiDoll

$\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}}$ 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

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

cough common core cough xD

33. myininaya

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

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

that is one the law of exponents

36. anonymous

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

ah... so would this work? $\large e^{\frac{1}{2}} \rightarrow e^{1}e^{-2}$

38. myininaya

no

39. myininaya

$e^1e^{-2}=e^{1-2}=e^{-1}=\frac{1}{e}$

40. UsukiDoll

oh xD now I remember

41. myininaya

have you considered getting rid of the square root part on bottom as I suggested earlier ?

42. UsukiDoll

e and the other exponent laws are different $n^{a-b} = \frac{n^a}{n^b}$

43. myininaya

n can be e

44. UsukiDoll

yeah $\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2}$

45. UsukiDoll

oh whoops yeah for something else not the exponent law thing

46. anonymous

maybe log.. can be applied?

47. myininaya

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

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

square root thing*

50. anonymous

My calculator says this will end up being 1/Sqrt[2E] and I would get a value of 0.428882

51. myininaya

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

are we on track?

53. myininaya

$y=\frac{\sqrt{2e}}{2e} \\ \text{ this is my preferred form } \text{ multiply } \sqrt{2e} \text{ on bottom and \top and simplify from } \\ \text{ there if you want } y=\frac{1}{\sqrt{2e}}$

54. dan815

sounds like a fun course :)

55. anonymous

lol, dan you would own this course.. :)

56. anonymous

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

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

ah I like that perspective.. thnx. myininaya, I will add that to the list..

59. UsukiDoll

oh I see.. after we got $e^\frac{1}{2}$ we changed it in radical form and rationalize!

60. dan815

|dw:1434788368933:dw|

61. myininaya

@hughfuve I still don't understand why you wanted to write it in this particular form

62. UsukiDoll

$\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{e}}$ $\frac{\sqrt{2}}{2\sqrt{e}}$ $\frac{\sqrt{2}}{2\sqrt{e}} \cdot \frac{\sqrt{e}}{\sqrt{e}}$ $\frac{\sqrt{2e}}{2e}$

63. dan815

u wanna graph it? it would be a horizontal line

64. myininaya

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

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

and do it by hand... there are no calculators allowed.

67. UsukiDoll

O_O!

68. dan815

u can use those series to approximate e

69. anonymous

well, that's just a single point.. on a graph, but it's a critical point.. it's the maxima

70. dan815

writing out the answer as y=1/sqrt(2e), is the nicest way imo

71. anonymous

this equation is the 1st derivative[0]

72. myininaya

Can you give the whole question?

73. anonymous

okay... 1 sec.. typing it in..

74. anonymous

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

first derivative will have something like y' = ....

76. UsukiDoll

oh this is another question...

77. anonymous

this is where this equation came from... I worked out the derivative.. and then put it in as x for f[x]

78. myininaya

so when you set f'=0 you got 1-2x^2=0 right?

79. anonymous

maybe I was completely on the wrong track though.

80. anonymous

argh.. I ballsed up there hey?

81. anonymous

82. anonymous

f'[x] = -2x^2 E^(-x^2) + E^(-x^2)

83. myininaya

right and you can factor the exp(-x^2) out

84. anonymous

0 = E^(-x^2) -2x^2 E^(-x^2)

85. anonymous

86. myininaya

$f'=e^{-x^2}(1-2x^2) \\ e^{-x^2} \neq 0 \text{ so we only have \to solve } 1-2x^2=0$

87. anonymous

I see...

88. anonymous

looking at the equation now and it's jumping out at me.

89. myininaya

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

but you still need to determine if they are max,min,neither

91. myininaya

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

2nd derivative? or is there a better way

93. myininaya

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

2nd derivative is concavity.

95. myininaya

looking at this graph can you tell me for what intervals y is positive and what intervals y is negative

96. anonymous

somewhere between +/- 1/2 and 1/1

97. anonymous

.5 .. .7 ... 1 ?

98. anonymous

sqrt 2 = about 1.4 ?

99. UsukiDoll

1/1 = 1 though

100. myininaya

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

thats why I was thinking ... 1/1 ... 1/sqrt(2) ... 1/2 somewhere in the middle.

102. anonymous

and came up with about .7

103. UsukiDoll

sqrt(2) = 1.41 1... 1/1.41 = 0.70 1/2 = 0.5

104. anonymous

or do we only need to know if positive or negative?

105. anonymous

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

perl suggested it earlier on a problem.

107. myininaya

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

so we have our parabola orientation. sweet.

109. myininaya

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

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

remember the f'=- means f decreasing and f'=+ means f increasing

112. anonymous

yes min in the -x max in the +x

113. myininaya

right so the min value is f(-1/sqrt(2)) and the max value is f(1/sqrt(2))

114. anonymous

/ \ 0 \ /

115. myininaya

so we know our graph looks something like this between -1/sqrt(2) and 1/sqrt(2) |dw:1434790667090:dw|

116. myininaya

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

and those points will be at.. y = f[ +/- 1/Sqrt[2] ]

118. myininaya

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

oh, actually..they gave me the points for that.. so we can skip that..

120. anonymous

its on a plot... two points near + / - (2,0)

121. myininaya

and I didn't mean to draw a 1 at those points by the way let me redo that

122. myininaya

|dw:1434790888537:dw|

123. myininaya

|dw:1434790920439:dw| you still need to finish the tail ends of the graphs where are they decreasing from do you think

124. myininaya

decreasing from and to *

125. anonymous

nice

126. anonymous

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

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

nice

129. myininaya

and remember sqrt(2e)/(2e) is the same number as 1/sqrt(2e) we just wrote it differently

130. anonymous

gotcha

131. myininaya

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

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

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

|dw:1434792046243:dw| yep so you would end up with something like this

135. anonymous

gotcha.. phew! exactly what I was thinking..

136. anonymous

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

137. anonymous

but this will get me a long way to solving the rest..

138. myininaya

good luck on the rest

139. myininaya

myininaya is probably going to have to go now unless you have anymore questions about this particular question?

140. anonymous

thank you thank you again.. I hope the rest of your night are fun problems.

141. anonymous

Im good

142. myininaya

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

143. myininaya

my whole night*

144. anonymous

lol.. oh wow.. well I appreciate your time very very much, you're a life saver.

145. myininaya

I wouldn't say wasted. I had fun. Played some video games and did some math problems. Fun fun.

146. myininaya

anyways goodnight(sort of )

147. anonymous

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

sigh.. going to be a long night. lol.

149. myininaya

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

150. UsukiDoll

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