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

This Question is Closed

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh wow what is this lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y = (Sqrt[2]/2) / E^((Sqrt[2]/2)^2) I think it's this.. \[y =\frac{ \frac{\sqrt{2}}{2} }{ E^{\frac{\sqrt{2}}{2}^2} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im supposed to be able to do this by hand... :/

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0OMG who assigns this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its the worst calculus course ever invented.. bill davis, horacio porta and Jerry Uhl

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0apparently a flipped curriculum.. where they explain nothing, and try to have the students work it out for themselves.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ffffffffffffffffffffffff

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[(\frac{\sqrt{2}}{2})^2\] I would start expanding this first.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0how is this Calculus? More like torture.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe \[ (\frac{2^.5} { 2 }) (\frac{2^.5}{ 2 })\] \[ \frac{(2^.5)(2^.5)} { 4 } \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ive been thinking the same thing.. lol

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ummm ... \[\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah.. is that true?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\LARGE e^{\frac{1}{2}}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0darn it ... I sort of forgot some of the e rules I know for \[\large e^{2+x} \rightarrow e^2e^x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I think I can move the 2 in the numerator down to the denominator

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I need to relook it up... I think it involves one of em .

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6hey so you are suppose to "simplify" the expression called y? and y is given as: \[y=\frac{\frac{\sqrt{2}}{2}}{e^{(\frac{\sqrt{2}}{2})^2}}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yeah and I already expanded the bottom

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0because that's an easy place to start

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I think we get y= 2^(1/2) / 2 E^(1/2)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0where did that 2 come from? ^ at the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw is there an easy way to enter these equations? that equation tool takes me forever.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6you can also write that as: \[y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6you can also chose to multiply the top and bottom by e^(1/2) to rationalize the denominator

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6well sorta rationalize anyways since e isn't rational :p

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}} \] 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}

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I cant believe these questions aren't all over open study.. I must be the only idiot in the world doing this course..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0cough common core cough xD

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6yes you can do this: \[y=\frac{\sqrt{2}}{2 e^\frac{1}{2}} \cdot \frac{e^\frac{1}{2}}{e^\frac{1}{2}}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0wasn'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? \[\large e^{2+x} \rightarrow e^2e^x \]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6that is one the law of exponents

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ah... so would this work? \[\large e^{\frac{1}{2}} \rightarrow e^{1}e^{2}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6\[e^1e^{2}=e^{12}=e^{1}=\frac{1}{e}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh xD now I remember

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6have you considered getting rid of the square root part on bottom as I suggested earlier ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0e and the other exponent laws are different \[n^{ab} = \frac{n^a}{n^b}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yeah \[\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh whoops yeah for something else not the exponent law thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe log.. can be applied?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6remember this: \[y=\frac{\sqrt{2}}{2 e^\frac{1}{2}} \cdot \frac{e^\frac{1}{2}}{e^\frac{1}{2}} \\ y= \frac{\sqrt{2} e^\frac{1}{2}}{2 e}=\frac{\sqrt{2 e}}{2e}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6anyways 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My calculator says this will end up being 1/Sqrt[2E] and I would get a value of 0.428882

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6ok 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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6\[y=\frac{\sqrt{2e}}{2e} \\ \text{ this is my preferred form } \text{ multiply } \sqrt{2e} \text{ on bottom and \top and simplify from } \\ \text{ there if you want } y=\frac{1}{\sqrt{2e}}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1sounds like a fun course :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, dan you would own this course.. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6example: \[\frac{\sqrt{u}}{u} \\ \frac{\sqrt{u}}{u} \cdot \frac{\sqrt{u}}{\sqrt{u}} =\frac{u}{u \sqrt{u}}=\frac{1}{\sqrt{u}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah I like that perspective.. thnx. myininaya, I will add that to the list..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh I see.. after we got \[e^\frac{1}{2}\] we changed it in radical form and rationalize!

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6@hughfuve I still don't understand why you wanted to write it in this particular form

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{e}} \] \[\frac{\sqrt{2}}{2\sqrt{e}}\] \[\frac{\sqrt{2}}{2\sqrt{e}} \cdot \frac{\sqrt{e}}{\sqrt{e}}\] \[\frac{\sqrt{2e}}{2e}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.1u wanna graph it? it would be a horizontal line

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6And 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 \[\frac{\sqrt{2e}}{2e} \text{ where } 0<\sqrt{2e}<2e \text{ dividing both sides by } 2e \text{ gives } \\ 0<\frac{\sqrt{2e}}{2e}<1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh Im just trying to get it to a place, where I can calculate a number , I need to arrive at 0.428882

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and do it by hand... there are no calculators allowed.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1u can use those series to approximate e

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, that's just a single point.. on a graph, but it's a critical point.. it's the maxima

dan815
 one year ago
Best ResponseYou've already chosen the best response.1writing out the answer as y=1/sqrt(2e), is the nicest way imo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this equation is the 1st derivative[0]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6Can you give the whole question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay... 1 sec.. typing it in..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Examine 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 ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0first derivative will have something like y' = ....

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh this is another question...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is where this equation came from... I worked out the derivative.. and then put it in as x for f[x]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6so when you set f'=0 you got 12x^2=0 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe I was completely on the wrong track though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0argh.. I ballsed up there hey?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f'[x] = 2x^2 E^(x^2) + E^(x^2)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6right and you can factor the exp(x^2) out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00 = E^(x^2) 2x^2 E^(x^2)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6\[f'=e^{x^2}(12x^2) \\ e^{x^2} \neq 0 \text{ so we only have \to solve } 12x^2=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looking at the equation now and it's jumping out at me.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6but looking at \[f(x)=xe^{x^2} \ \\ f(\frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}e^{(\frac{1}{\sqrt{2}})^2}=\frac{1}{\sqrt{2}}e^{\frac{1}{2}}=\frac{1}{\sqrt{2} e^\frac{1}{2}}=\frac{1}{\sqrt{2e}} \\ \text{ you already got this \above } \\ \text{ now you can find the other value } f(\frac{1}{\sqrt{2}})\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6but you still need to determine if they are max,min,neither

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6e^(x^2) is positive for all x (12x^2) is a parabola; it is what will tell us if the derivative is positive or negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02nd derivative? or is there a better way

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6this is a rough graph of y=12x^2 dw:1434789467495:dw we are just using this graph to determine when f' is positive or negative you know where f' is: \[f'(x)=e^{x^2}(12x^2)\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.02nd derivative is concavity.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6looking at this graph can you tell me for what intervals y is positive and what intervals y is negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0somewhere between +/ 1/2 and 1/1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sqrt 2 = about 1.4 ?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6I'm trouble understanding what you are saying ... umm... in the picture I drew of the parabola y=12x^2 y is positive when the graph is above the xaxis it is positive on (1/sqrt(2),1/sqrt(2)) y is negative when the graph is below the xaxis it is negative on (inf,1/sqrt(2)) U (1/sqrt(2),inf)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats why I was thinking ... 1/1 ... 1/sqrt(2) ... 1/2 somewhere in the middle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and came up with about .7

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0sqrt(2) = 1.41 1... 1/1.41 = 0.70 1/2 = 0.5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or do we only need to know if positive or negative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on 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 ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0perl suggested it earlier on a problem.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6this means: \[f'(x)=e^{x^2}(12x^2) \\ f' \text{ is positive on } (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} ) \text{ which means } f \text{ is increasing there } \\ f' \text{ is negative on } (\infty, \frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \text{ which means } f \\ \text{ is decreasing there }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we have our parabola orientation. sweet.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6anyways 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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6so 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?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6remember the f'= means f decreasing and f'=+ means f increasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes min in the x max in the +x

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6right so the min value is f(1/sqrt(2)) and the max value is f(1/sqrt(2))

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6so we know our graph looks something like this between 1/sqrt(2) and 1/sqrt(2) dw:1434790667090:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6now 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and those points will be at.. y = f[ +/ 1/Sqrt[2] ]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6I ask you to consider the following limits for end behavior : \[\lim_{x \rightarrow \infty}f(x) \text{ and } \lim_{x \rightarrow \infty}f(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, actually..they gave me the points for that.. so we can skip that..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its on a plot... two points near + /  (2,0)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6and I didn't mean to draw a 1 at those points by the way let me redo that

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6dw:1434790888537:dw

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6dw:1434790920439:dw you still need to finish the tail ends of the graphs where are they decreasing from do you think

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6decreasing from and to *

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you wanted now to work out the square root of 2E by hand.. is there a trick? other than trial and error and squeezing ?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6you 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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6and remember sqrt(2e)/(2e) is the same number as 1/sqrt(2e) we just wrote it differently

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6summary: \[f(x)=xe^{x^2} \\ f'(x)=e^{x^2}+x(2x)e^{x^2} \\ f'(x)=e^{x^2}(12x^2) \\ \text{ we know } e^{x^2} >0 \text{ for all } x \\ 12x^2=0 \text{ when } x=\pm \frac{1}{\sqrt{2}} \text{ or if you prefer \to write as } x=\pm \frac{\sqrt{2}}{2} \\ \text now f(\pm \frac{\sqrt{2}}{2})=\pm \frac{1}{\sqrt{2e}} \text{ or if you prefer } \pm \frac{\sqrt{2e}}{2e} \\ \text{ also looking at } f' \\ \text{ we have } f'>0 \text{ when } x \in (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) \text{ so } f \text{ is increasing on } (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) \\ \text{ we have } f'<0 \text{ when } x \in (\infty,\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \\ \text{ so } f \text{ is decreasing on } (\infty,\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \\ \text{ so this tells us our max } f(\frac{\sqrt{2}}{2})=\frac{1}{\sqrt{2e}} \text{ and our min is } f(\frac{\sqrt{2}}{2})=\frac{1}{\sqrt{2e}} \] 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6you 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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6dw:1434792046243:dw yep so you would end up with something like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0gotcha.. phew! exactly what I was thinking..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0man, I wish every problem wasn't like this. I got 11 more on this assignment.. just like this one.. lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but this will get me a long way to solving the rest..

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6good luck on the rest

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6myininaya is probably going to have to go now unless you have anymore questions about this particular question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you thank you again.. I hope the rest of your night are fun problems.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6I actually wasted my whole. I should have been in bed like hours ago. It is 4:25 AM here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol.. oh wow.. well I appreciate your time very very much, you're a life saver.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6I wouldn't say wasted. I had fun. Played some video games and did some math problems. Fun fun.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6anyways goodnight(sort of )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here's a sneak peak at my next challenge.. please dont answer ... To Plot this by hand ... f[x] = (x^2  2 Log[x])/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sigh.. going to be a long night. lol.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.6http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtI.aspx This site maybe helpful too. Not sure if you ever heard of it.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0at 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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