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anonymous
 one year ago
Whats the trick to solve for x with this equation?
0 = (2x2x^3) E^(1x^2)
x = ???
anonymous
 one year ago
Whats the trick to solve for x with this equation? 0 = (2x2x^3) E^(1x^2) x = ???

This Question is Closed

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1what is this (2x2x^3) E^(1x^2)

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3\[(2x2x^3)e ^{(1x^2)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its a derivative of f[x] = x^2 E^(x^2 + 1) apparently I have to examine the derivative and find the zeros, without a calculator, to plot a chart.. I can see just looking at the equations that 0,1,1 will give me the zeros.. but I can't explain why that would be true..

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1why would you need a calculator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, whats latex ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why a calc, coz I'm clueless of course..

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1you probably need to use log

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I was worried about that..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To get rid of that E^ ?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3No, 1 and 1 are not zeros.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0had me there for minute.. thought I sent the wrong equation

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I am confused how you can "see" the values without solving it.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3He may have looked at the graphical representation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It just looks obvious to me.. I usually plug 1 and 0 in to get an idea of the function and if 1 zeros, 1 usually does too when squares are involved.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1then use that as your mathematical solution "it is obvious"

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Nin, don't be so rude.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1@UnkleRhaukus @UsukiDoll @rational

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3if you want to find the zeroes of the function: \[(2x2x^3)e ^{1x^2}=0\] is clearly and application of the hankelian preoperty, wich means: \[2x2x^3=0\] \[e ^{1x^2}= 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats a cool one.. first time I've heard of that.. but it makes sense yes.. if a*b = 0 then a or b = 0

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3yes, now all you have to do is find the zeroes of those smaller and simpler parts of the function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0silly question ... does that also apply for division?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Not really, I think you mean: \[\frac{ a }{ b }=0 <=>a=0,b=0 \] Thing is that does not work, because the expression a/b has the followin condition: \[\frac{ a }{ b }=0 <=> a \in \mathbb{R}, b \neq 0\] because you can't have a zero in the denominator.

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1hint : \[e^{something}\neq0\]

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Do you mean that it does no have zeroes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah amoodarya, thnx.. yea E^0 = 1

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1can you prove that \(e^{1x^2} = 0 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(2x2x^3) is all we need though right? because if that is 0 the whole equation is 0

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3oh no, wait, log 0 is not possible.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I know it is not possible, but you said you can

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, I tried to log both sides earlier.. when one side was 0, can you do that even temporarily ?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, thought there would be a constant.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1this is beyond my skill set

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if the zero must be in the \[0 =(2x  2x^3)\] then \[2x = 2x  2x^3 2x\] \[2x =  2x^3 \] \[2x/2 =  2x^3 /2 \] \[x = x^3 \] \[x/x = x^3/x \] \[1 = x^2 \] \[1/x = x^2/x \] x = 1/x Does this look right?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Well, not really, just because : \[e ^{f(x)}=0\] Will always have positive images of x, because it is a lil' more complex than the expression: \[f:f(x)=e ^{x}\] therefore, the only part of the function that'll determine the zeroes will be: \[(2x2x^3)\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I am interested to learn this

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3To find the zeroes of the expression: \[2x2x^3=0\] You take common factor 2x: \[2x(1x^2)\] And again, by the hankelian propety: \[2x=0 \] \[1x^2 =0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the solution might also be here.. x = x^3 x = x x x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ummm.... is this solve for x or find the derivative ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3Alright... I'm going to use what @Owlcoffee suggested earlier.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[(2x2x^3)e ^{1x^2}=0\] so we're going to solve for x... more like find all x's that satisfy the equation... this feels like find the equilibrium or something. \[(2x2x^3)=0\] \[e ^{1x^2}=0\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3for the first equation we have a 2x in common \[(2x2x^3) \rightarrow 2x(1x^2)\] factoring again yields \[2x(1+x)(1x)=0\] just through analysis and looking at it x = 0,1,1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[e ^{1x^2}=0\] if I take the ln on both sides because e^x and ln x are inverses of each other, that e should go away ... the problem is \[e^{\ln(1x^2)}=\ln 0 \] 1x^2 = ln 0 and ln 0 is undefined >:/

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3in fact ln (negative any number) is also undefined.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1the easy part is always the best part, yes?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Nin, what the hell? Usuki made a great job explaining it very very simple.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3O____________________________________o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it not be okay though if \[E^(1x^2) = 1\] because the zeros on the other term would satisfy what we need

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3do you mean \[e^{1x^2} = 1\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah I think if x = 1,0,1 that equation = 1 , and that's okay.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I still do not get E what is E

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the exponential constant = E

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1so it is not Euler's constant?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3well let's supposed we are given that equation instead \[e^{1x^2} = 1\] so taking the ln on both sides yields \[e^{\ln(1x^2)} = \ln 1\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3? What does Euler's Method have to do with this? That's differential equations

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1you're dazzling as always, @UsukiDoll did you get your degree yet?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3that question has nothing to do with what we are trying to solve. Please delete it.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3If you know so much, Nin, why aren't you helping?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ok so e is the mathematical exponent.... it's also Euler's number... and it's default value is 2.71... but enough with that.. let's go back to what we're doing...this is going off topic and it's not helping hughfuve

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so assuming we have \[e^{\ln(1x^2)} = \ln 1\] since ln 1 > 0 \[1x^2=0\] \[(1+x)(1x) = 0\] \[x=1,1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is it a valid technique to solve for that first half of the equation... (2x  2x^3) and then take the 3 zeros that you found.. and plug them into the second part of the equation. Or would that be frowned upon?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I'm trying to think if this is going to work... if I plug in x = 1 on the original equation... I will have \[e^{\ln 0} = 0\] I know ln 0 is undefined.. but \[e^x \] and \[\ln x \] are inverses of each other so I can use that to make the equation hold true. wait let me see that suggestion..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3our zeros from the first equation x = 0,1,1 x=1 is going to be a problem if I put it in the second equation

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[e ^{1x^2}=0\] let x = 1 \[e ^{1(1)^2}=0\] \[e ^{11}=0\] \[e ^{0}=0\] \[1 \neq 0\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3x = 1 will produce the same issue to because (1)^2  > (1)(1) = 1 Same thing will occur

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Thing is, if you plug the zeroes you found, say for instance, x=0: \[f(0)=(2(0)2(0)^3)e ^{10^2}\] you'll end up having: \[f(0)=0.e^1\] which is just : \[f(0)=0\] And that'll happen for 1, and 1. it will make the whole function a "0", but it will not make e^(1x^2) equal zero.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3and we can't use x = 0 either you will end up with \[e^1 = 0 \rightarrow 2.71 \neq 0 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but does E^(1x^2) have to equal zero? if the full equation is \[0 == (2x2x^3) (E^1x^2)\] then even if E^(n) = 1 or E, is our equation not still a valid 0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops on my equation.. its supposed to have everything after E in the exponent.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the zero's we found for x =0,1,1 is not going to make \[e^{(1x^2)}=0\] true for 2 reasons. you will either have \[e^0 \neq 0 \rightarrow 1 \neq 0\] or \[e \neq 0 \rightarrow 2.71 \neq 0 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but I think if you combine those two together... the left side of the equation the one with the 2x should make it go 0... yeah let's try that MAHAHAHAH!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess I'm not understanding why we are searching for 0 on that portion. If A * B = 0 right, then only A OR B needs to be zero.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Thing is the understanding you have for the function: \[f:f(x)= e^x\] if you look at the graph, you'll see that it has no zeroes.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[0 =(2x2x^3) (e^{(1x^2)})\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so if I substituted x = 0 \[0 =(2(0)2(0)^3) (e^{(1(0)^2)})\] \[0 =(2(0)2(0)) (e^{(10)})\] \[0 =(00)) (e^{(1)})\] \[0=(0)(e)\] e is approximately 2.71 0=(0)(2.71) 0=0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3it's like we need that \[(2x2x^3)\] portion to help us get rid of the e ... otherwise we're stuck because x = 1 yields (22) > (0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0totally, cant escape it hey

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I think .. splitting the problem up is only necessary to find the zeros... but we need the whole problem together and have x = 0,1,1 in order for that equation to hold true. If we keep the split up version... something isn't going to cancel and will lead to something false.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think that will be an adequate solution..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0factor out the first part of the problem, and claim the 2nd term as basically irrelevant.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yeah.. because by setting the second equation to 0 , you will run into problems.. because ln 0 is undefined. If that's the case, just stick with the first part of the equation.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I've done this before when I had to find the equilibrium for Mathematical biology... if we come across something that's undefined...ignore it and focus on the problem that will give us the 0's .. in this case it was the first part of the equation which gave us x =0,1, 1.. plugging those x values back into the whole equation, will make your equation 0=0 which is true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I guess I should have mentioned, I just realized it might be important to know, there is a fixed domain too by the way.. it's +2 .. 2 so we dont need to worry about  infinity

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ok no problem.. the domain lies in the xaxis

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434530843604:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434530877368:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I am thinking, within the domain there is no way for E^anything to become 0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3well e itself is 2.71 and e^0 is 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I can claim then as there is no way for E^(1x^2) within the domain (2,2) to become 0, then the zero is to be found in the other part of the equation.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yes the zeros are found in the other part of the equation .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks Usuki.. you're awesome. and thanks to everyone who came to help.. much appreciated.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3and if we have x = 0 we end up with e which is 2.711 and for x =1,1 we have e^0 which is 1... I am unable to have that equation go to 0 .

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I think I did way more than the Qualified Helper... if you've paid for this question... you got ripped off :(

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3actually it's @Owlcoffee and me doing all the work

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but still if I had to pay for qualified help, I would expect to have extraordinary result and expert explanations, not chit chat :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, well, you saved the day then :)

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, i wish i could've helped better :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0true, sometimes the help is not so helpful.. it costs. 10 a question I think

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3sorry 10 owlbucks is $2 I think... you didn't pay that much did you? I notice your question was highlighted in yellow

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3@Owlcoffee you did a good job too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0owlcoffee, yes you did great.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but my main concern is whether or not you have already paid for qualified help... because if you did, I would ask for a refund and reevaluation of the qualified helper in question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually. that email address they say you get a refund on.. doesnt work.. it bounces.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3:(... but that highlighted yellow question meant that you've paid for qualified help right?

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3You paid for a qualified helper!?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3his question was highlighted in yellow... unusual since free questions are in grey

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, I just figured it was too late at night and bad timing.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3=( get a refund...pm a moderator and explained what happened. You've been ripped off

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I pay for most my questions.. they're not always handled too well. but often I get a very conscientious helper..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3not for this one... this qh was chit chatting and barely did anything.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3I believe qualified helpers should be evaluated by the help given and not smartscore. Nincoops is sadly, not a "helper". Glad you came @UsukiDoll I was about to just leave this question here.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3thanks :) on top of that nin confessed to not knowing and had the nerve to tag me and 2 99ers who I am fans with. Like, if you're a qualified helper... you should know what the question is already talking about and guide the person through the whole process.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3qualified helper < know what's going on already...and be able to explain and clarify, not omg I have no idea what'a going on ... let me just tag someone else. That's not being a qualified helper. That's a qualified FRAUDULENT helper.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3so, regardless of that, if you ever need help you can tag me or @UsukiDoll and we'll help you as soon as we can. @hughfuve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0guys thats's awesome thank you, I kind of dont want to rock the boat around here.. I really need the help right now.. I'm like 4 chapters into a 15 chapter course and I'm already hitting the halfway timer.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3it depends on the question...there are some topics that I've forgot though since it has been a very long time .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0got about 5 weeks to get through the rest of it.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3is this a summer school course?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I remember having to retake college algebra during the summer.. it was crazy intense 6 weeks, but I already knew the material so it wasn't that bad. I had to retake because the professor I had during the spring semester was tough as nails.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its a disaster of a curriculum for calculus using mathematica, but someone has butchered it up, and everything is out of order..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0at least my professor is pretty cool.. having an retriceof a teacher would be awful

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3is it calculus 1,2,3, or 4?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it';s an entry level its 234 business calc

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3oo. Oh I'm horrible at Business Calculus . I used to be a business major.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3It's strange... Business Calculus makes me look like an idiot, but regular Calculus is easy.. I passed all 4 sections with above average grades.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but its just poorly explained.. if I didnt have calc videos, I'd be in bad trouble. they just throw the equations at you.. say.. this is what mathetmatica did.. now explain it..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3Mathematica is a computer program.. I had to use it for Calculus II

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like they're throwing derivatives at you.. in chapter 3, and asking you what they are, when they dont even introduce them until chapter 4.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmmm that could be the definition of the derivative... I had to use that long beast before using the real derivative techniques.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{f(x+h)f(x)}{h}\]

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3\[\lim_{x \rightarrow a}\frac{ f(x)f(a) }{ xa }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, that part was ugly.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, man was that ugly

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3that looks like mean value theorem xD or the rate of change formula

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3it is the definition of derivative. just a variation of it, there are 3, I think.

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

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.3By definition, the derivative gives you the slope of the tangent line on any point of a function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess thats uski's difference quotient up there too = slope

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay.. time to ask another Q ...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I might be away from keyboard. I have to do something.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0starting a new thread.. all good.
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