anonymous
  • anonymous
Whats the trick to solve for x with this equation? 0 = (2x-2x^3) E^(1-x^2) x = ???
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
nincompoop
  • nincompoop
what is this (2x-2x^3) E^(1-x^2)
nincompoop
  • nincompoop
use latex
Owlcoffee
  • Owlcoffee
\[(2x-2x^3)e ^{(1-x^2)}\]

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anonymous
  • anonymous
its a derivative of f[x] = x^2 E^(-x^2‚Äč + 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
  • nincompoop
why would you need a calculator?
anonymous
  • anonymous
sorry, whats latex ?
anonymous
  • anonymous
why a calc, coz I'm clueless of course..
nincompoop
  • nincompoop
you probably need to use log
anonymous
  • anonymous
Yeah I was worried about that..
anonymous
  • anonymous
To get rid of that E^ ?
Owlcoffee
  • Owlcoffee
No, 1 and -1 are not zeros.
nincompoop
  • nincompoop
they are zeroes
anonymous
  • anonymous
had me there for minute.. thought I sent the wrong equation
nincompoop
  • nincompoop
I am confused how you can "see" the values without solving it.
Owlcoffee
  • Owlcoffee
He may have looked at the graphical representation.
anonymous
  • anonymous
It 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
  • nincompoop
then use that as your mathematical solution "it is obvious"
Owlcoffee
  • Owlcoffee
Nin, don't be so rude.
nincompoop
  • nincompoop
@UnkleRhaukus @UsukiDoll @rational
Owlcoffee
  • Owlcoffee
if you want to find the zeroes of the function: \[(2x-2x^3)e ^{1-x^2}=0\] is clearly and application of the hankelian preoperty, wich means: \[2x-2x^3=0\] \[e ^{1-x^2}= 0\]
anonymous
  • anonymous
thats 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
  • Owlcoffee
yes, now all you have to do is find the zeroes of those smaller and simpler parts of the function.
anonymous
  • anonymous
silly question ... does that also apply for division?
Owlcoffee
  • Owlcoffee
Not 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
  • amoodarya
hint : \[e^{something}\neq0\]
Owlcoffee
  • Owlcoffee
Do you mean that it does no have zeroes?
anonymous
  • anonymous
ah amoodarya, thnx.. yea E^0 = 1
nincompoop
  • nincompoop
can you prove that \(e^{1-x^2} = 0 \)
Owlcoffee
  • Owlcoffee
I can.
anonymous
  • anonymous
(2x-2x^3) is all we need though right? because if that is 0 the whole equation is 0
Owlcoffee
  • Owlcoffee
oh no, wait, log 0 is not possible.
nincompoop
  • nincompoop
I know it is not possible, but you said you can
anonymous
  • anonymous
yeah, I tried to log both sides earlier.. when one side was 0, can you do that even temporarily ?
Owlcoffee
  • Owlcoffee
Sorry, thought there would be a constant.
nincompoop
  • nincompoop
this is beyond my skill set
anonymous
  • anonymous
so 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?
anonymous
  • anonymous
lol, it cant be
Owlcoffee
  • Owlcoffee
Well, 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: \[(2x-2x^3)\]
nincompoop
  • nincompoop
n0
nincompoop
  • nincompoop
I am interested to learn this
Owlcoffee
  • Owlcoffee
To find the zeroes of the expression: \[2x-2x^3=0\] You take common factor 2x: \[2x(1-x^2)\] And again, by the hankelian propety: \[2x=0 \] \[1-x^2 =0\]
anonymous
  • anonymous
I think the solution might also be here.. x = x^3 x = x x x
nincompoop
  • nincompoop
uh...
anonymous
  • anonymous
lol, maybe not..
nincompoop
  • nincompoop
stop trolling
UsukiDoll
  • UsukiDoll
ummm.... is this solve for x or find the derivative ?
anonymous
  • anonymous
solve for x
UsukiDoll
  • UsukiDoll
Alright... I'm going to use what @Owlcoffee suggested earlier.
UsukiDoll
  • UsukiDoll
\[(2x-2x^3)e ^{1-x^2}=0\] 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. \[(2x-2x^3)=0\] \[e ^{1-x^2}=0\]
nincompoop
  • nincompoop
good luck
UsukiDoll
  • UsukiDoll
for the first equation we have a 2x in common \[(2x-2x^3) \rightarrow 2x(1-x^2)\] factoring again yields \[2x(1+x)(1-x)=0\] just through analysis and looking at it x = 0,1,-1
UsukiDoll
  • UsukiDoll
\[e ^{1-x^2}=0\] 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 \[e^{\ln(1-x^2)}=\ln 0 \] 1-x^2 = ln 0 and ln 0 is undefined >:/
UsukiDoll
  • UsukiDoll
in fact ln (negative any number) is also undefined.
nincompoop
  • nincompoop
the easy part is always the best part, yes?
Owlcoffee
  • Owlcoffee
Nin, what the hell? Usuki made a great job explaining it very very simple.
UsukiDoll
  • UsukiDoll
O____________________________________o
anonymous
  • anonymous
would it not be okay though if \[E^(1-x^2) = 1\] because the zeros on the other term would satisfy what we need
UsukiDoll
  • UsukiDoll
do you mean \[e^{1-x^2} = 1\] ?
anonymous
  • anonymous
yeah I think if x = 1,0,-1 that equation = 1 , and that's okay.
anonymous
  • anonymous
or E
nincompoop
  • nincompoop
I still do not get E what is E
anonymous
  • anonymous
the exponential constant = E
nincompoop
  • nincompoop
so it is not Euler's constant?
UsukiDoll
  • UsukiDoll
well let's supposed we are given that equation instead \[e^{1-x^2} = 1\] so taking the ln on both sides yields \[e^{\ln(1-x^2)} = \ln 1\]
anonymous
  • anonymous
sorry, yes Eulers
UsukiDoll
  • UsukiDoll
? What does Euler's Method have to do with this? That's differential equations
nincompoop
  • nincompoop
you're dazzling as always, @UsukiDoll did you get your degree yet?
UsukiDoll
  • UsukiDoll
that question has nothing to do with what we are trying to solve. Please delete it.
nincompoop
  • nincompoop
:)
Owlcoffee
  • Owlcoffee
If you know so much, Nin, why aren't you helping?
UsukiDoll
  • UsukiDoll
ok 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
  • UsukiDoll
so assuming we have \[e^{\ln(1-x^2)} = \ln 1\] since ln 1 -> 0 \[1-x^2=0\] \[(1+x)(1-x) = 0\] \[x=1,-1\]
anonymous
  • anonymous
Is 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
  • UsukiDoll
I'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
  • UsukiDoll
our 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
  • UsukiDoll
\[e ^{1-x^2}=0\] let x = 1 \[e ^{1-(1)^2}=0\] \[e ^{1-1}=0\] \[e ^{0}=0\] \[1 \neq 0\]
UsukiDoll
  • UsukiDoll
x = -1 will produce the same issue to because (-1)^2 - > (-1)(-1) = 1 Same thing will occur
Owlcoffee
  • Owlcoffee
Thing is, if you plug the zeroes you found, say for instance, x=0: \[f(0)=(2(0)-2(0)^3)e ^{1-0^2}\] 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^(1-x^2) equal zero.
UsukiDoll
  • UsukiDoll
and we can't use x = 0 either you will end up with \[e^1 = 0 \rightarrow 2.71 \neq 0 \]
anonymous
  • anonymous
but does E^(1-x^2) have to equal zero? if the full equation is \[0 == (2x-2x^3) (E^1-x^2)\] then even if E^(n) = 1 or E, is our equation not still a valid 0?
anonymous
  • anonymous
oops on my equation.. its supposed to have everything after E in the exponent.
UsukiDoll
  • UsukiDoll
the zero's we found for x =0,-1,1 is not going to make \[e^{(1-x^2)}=0\] 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
  • UsukiDoll
but 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
  • anonymous
I 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
  • Owlcoffee
Thing 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
  • UsukiDoll
\[0 =(2x-2x^3) (e^{(1-x^2)})\]
UsukiDoll
  • UsukiDoll
so if I substituted x = 0 \[0 =(2(0)-2(0)^3) (e^{(1-(0)^2)})\] \[0 =(2(0)-2(0)) (e^{(1-0)})\] \[0 =(0-0)) (e^{(1)})\] \[0=(0)(e)\] e is approximately 2.71 0=(0)(2.71) 0=0
UsukiDoll
  • UsukiDoll
it's like we need that \[(2x-2x^3)\] portion to help us get rid of the e ... otherwise we're stuck because x = 1 yields (2-2) -> (0)
anonymous
  • anonymous
totally, cant escape it hey
UsukiDoll
  • UsukiDoll
I 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
  • anonymous
I think that will be an adequate solution..
anonymous
  • anonymous
factor out the first part of the problem, and claim the 2nd term as basically irrelevant.
UsukiDoll
  • UsukiDoll
yeah.. 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
  • UsukiDoll
I'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
  • anonymous
Sorry, 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
  • UsukiDoll
ok no problem.. the domain lies in the x-axis
UsukiDoll
  • UsukiDoll
|dw:1434530843604:dw|
UsukiDoll
  • UsukiDoll
|dw:1434530877368:dw|
anonymous
  • anonymous
So I am thinking, within the domain there is no way for E^anything to become 0
UsukiDoll
  • UsukiDoll
well e itself is 2.71 and e^0 is 1
anonymous
  • anonymous
So I can claim then as there is no way for E^(1-x^2) within the domain (-2,2) to become 0, then the zero is to be found in the other part of the equation.
UsukiDoll
  • UsukiDoll
yes the zeros are found in the other part of the equation .
anonymous
  • anonymous
Thanks Usuki.. you're awesome. and thanks to everyone who came to help.. much appreciated.
UsukiDoll
  • UsukiDoll
and 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
  • UsukiDoll
I think I did way more than the Qualified Helper... if you've paid for this question... you got ripped off :(
UsukiDoll
  • UsukiDoll
actually it's @Owlcoffee and me doing all the work
UsukiDoll
  • UsukiDoll
but still if I had to pay for qualified help, I would expect to have extraordinary result and expert explanations, not chit chat :/
anonymous
  • anonymous
lol, well, you saved the day then :)
Owlcoffee
  • Owlcoffee
Sorry, i wish i could've helped better :(
anonymous
  • anonymous
true, sometimes the help is not so helpful.. it costs. 10 a question I think
UsukiDoll
  • UsukiDoll
sorry 10 owlbucks is $2 I think... you didn't pay that much did you? I notice your question was highlighted in yellow
UsukiDoll
  • UsukiDoll
@Owlcoffee you did a good job too.
anonymous
  • anonymous
owlcoffee, yes you did great.
UsukiDoll
  • UsukiDoll
but 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 re-evaluation of the qualified helper in question
anonymous
  • anonymous
actually. that email address they say you get a refund on.. doesnt work.. it bounces.
UsukiDoll
  • UsukiDoll
:(... but that highlighted yellow question meant that you've paid for qualified help right?
Owlcoffee
  • Owlcoffee
You paid for a qualified helper!?
UsukiDoll
  • UsukiDoll
his question was highlighted in yellow... unusual since free questions are in grey
anonymous
  • anonymous
yes, I just figured it was too late at night and bad timing.
UsukiDoll
  • UsukiDoll
=( get a refund...pm a moderator and explained what happened. You've been ripped off
anonymous
  • anonymous
I pay for most my questions.. they're not always handled too well. but often I get a very conscientious helper..
UsukiDoll
  • UsukiDoll
not for this one... this qh was chit chatting and barely did anything.
UsukiDoll
  • UsukiDoll
right @Owlcoffee ?
Owlcoffee
  • Owlcoffee
I 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
  • UsukiDoll
thanks :) 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
  • UsukiDoll
qualified 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
  • Owlcoffee
so, 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
  • anonymous
guys 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
  • UsukiDoll
it depends on the question...there are some topics that I've forgot though since it has been a very long time .
anonymous
  • anonymous
got about 5 weeks to get through the rest of it.
UsukiDoll
  • UsukiDoll
is this a summer school course?
anonymous
  • anonymous
yeah
UsukiDoll
  • UsukiDoll
I remember having to re-take 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 re-take because the professor I had during the spring semester was tough as nails.
anonymous
  • anonymous
its a disaster of a curriculum for calculus using mathematica, but someone has butchered it up, and everything is out of order..
anonymous
  • anonymous
at least my professor is pretty cool.. having an retriceof a teacher would be awful
UsukiDoll
  • UsukiDoll
is it calculus 1,2,3, or 4?
anonymous
  • anonymous
I think it';s an entry level its 234 business calc
UsukiDoll
  • UsukiDoll
o-------o. Oh I'm horrible at Business Calculus . I used to be a business major.
UsukiDoll
  • UsukiDoll
It'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
  • anonymous
but 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
  • UsukiDoll
Mathematica is a computer program.. I had to use it for Calculus II
anonymous
  • anonymous
like 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
  • UsukiDoll
hmmm that could be the definition of the derivative... I had to use that long beast before using the real derivative techniques.
UsukiDoll
  • UsukiDoll
\[\frac{f(x+h)-f(x)}{h}\]
Owlcoffee
  • Owlcoffee
\[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]
anonymous
  • anonymous
yes, that part was ugly.
anonymous
  • anonymous
lol, man was that ugly
UsukiDoll
  • UsukiDoll
that looks like mean value theorem xD or the rate of change formula
Owlcoffee
  • Owlcoffee
it is the definition of derivative. just a variation of it, there are 3, I think.
anonymous
  • anonymous
is the other the slope formula?
Owlcoffee
  • Owlcoffee
By definition, the derivative gives you the slope of the tangent line on any point of a function.
anonymous
  • anonymous
I guess thats uski's difference quotient up there too = slope
anonymous
  • anonymous
okay.. time to ask another Q ...
UsukiDoll
  • UsukiDoll
I might be away from keyboard. I have to do something.
anonymous
  • anonymous
starting a new thread.. all good.

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