Question

Whats the trick to solve for x with this equation?

0 = (2x-2x^3) E^(1-x^2)
x = ???

Answer 1

what is this (2x-2x^3) E^(1-x^2)

Answer 2

use latex

Answer 3

\[(2x-2x^3)e ^{(1-x^2)}\]

Answer 4

its 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..

Answer 5

why would you need a calculator?

Answer 6

sorry, whats latex ?

Answer 7

why a calc, coz I'm clueless of course..

Answer 8

you probably need to use log

Answer 9

Yeah I was worried about that..

Answer 10

To get rid of that E^ ?

Answer 11

No, 1 and -1 are not zeros.

Answer 12

they are zeroes

Answer 13

had me there for minute.. thought I sent the wrong equation

Answer 14

I am confused how you can "see" the values without solving it.

Answer 15

He may have looked at the graphical representation.

Answer 16

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.

Answer 17

then use that as your mathematical solution
"it is obvious"

Answer 18

Nin, don't be so rude.

Answer 19

@UnkleRhaukus @UsukiDoll @rational

Answer 20

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:
\[2x-2x^3=0\]
\[e ^{1-x^2}= 0\]

Answer 21

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

Answer 22

yes, now all you have to do is find the zeroes of those smaller and simpler parts of the function.

Answer 23

silly question ... does that also apply for division?

Answer 24

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.

Answer 25

hint :
\[e^{something}\neq0\]

Answer 26

Do you mean that it does no have zeroes?

Answer 27

ah amoodarya, thnx.. yea E^0 = 1

Answer 28

can you prove that \(e^{1-x^2} = 0 \)

Answer 29

I can.

Answer 30

(2x-2x^3) is all we need though right? because if that is 0 the whole equation is 0

Answer 31

oh no, wait, log 0 is not possible.

Answer 32

I know it is not possible, but you said you can

Answer 33

yeah, I tried to log both sides earlier.. when one side was 0, can you do that even temporarily ?

Answer 34

Sorry, thought there would be a constant.

Answer 35

this is beyond my skill set

Answer 36

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?

Answer 37

lol, it cant be

Answer 38

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:
\[(2x-2x^3)\]

Answer 39

n0

Answer 40

I am interested to learn this

Answer 41

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:
\[2x=0 \]
\[1-x^2 =0\]

Answer 42

I think the solution might also be here..
x = x^3
x = x x x

Answer 43

uh...

Answer 44

lol, maybe not..

Answer 45

stop trolling

Answer 46

ummm.... is this solve for x or find the derivative ?

Answer 47

solve for x

Answer 48

Alright... I'm going to use what @Owlcoffee suggested earlier.

Answer 49

\[(2x-2x^3)e ^{1-x^2}=0\]
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.

\[(2x-2x^3)=0\]

\[e ^{1-x^2}=0\]

Answer 50

good luck

Answer 51

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

Answer 52

\[e ^{1-x^2}=0\]
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 >:/

Answer 53

in fact ln (negative any number) is also undefined.

Answer 54

the easy part is always the best part, yes?

Answer 55

Nin, what the hell?
Usuki made a great job explaining it very very simple.

Answer 56

O____________________________________o

Answer 57

would it not be okay though if
\[E^(1-x^2) = 1\]
because the zeros on the other term would satisfy what we need

Answer 58

do you mean \[e^{1-x^2} = 1\] ?

Answer 59

yeah I think if x = 1,0,-1 that equation = 1 , and that's okay.

Answer 60

or E

Answer 61

I still do not get E
what is E

Answer 62

the exponential constant = E

Answer 63

so it is not Euler's constant?

Answer 64

well let's supposed we are given that equation instead
\[e^{1-x^2} = 1\]
so taking the ln on both sides yields

\[e^{\ln(1-x^2)} = \ln 1\]

Answer 65

sorry, yes Eulers

Answer 66

? What does Euler's Method have to do with this? That's differential equations

Answer 67

you're dazzling as always, @UsukiDoll did you get your degree yet?

Answer 68

that question has nothing to do with what we are trying to solve. Please delete it.

Answer 69

:)

Answer 70

If you know so much, Nin, why aren't you helping?

Answer 71

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

Answer 72

so assuming we have
\[e^{\ln(1-x^2)} = \ln 1\]
since ln 1 -> 0
\[1-x^2=0\]
\[(1+x)(1-x) = 0\]
\[x=1,-1\]

Answer 73

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?

Answer 74

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..

Answer 75

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

Answer 76

\[e ^{1-x^2}=0\]
let x = 1
\[e ^{1-(1)^2}=0\]
\[e ^{1-1}=0\]
\[e ^{0}=0\]
\[1 \neq 0\]

Answer 77

x = -1 will produce the same issue to because (-1)^2 - > (-1)(-1) = 1
Same thing will occur

Answer 78

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.

Answer 79

and we can't use x = 0 either
you will end up with
\[e^1 = 0 \rightarrow 2.71 \neq 0 \]

Answer 80

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?

Answer 81

oops on my equation.. its supposed to have everything after E in the exponent.

Answer 82

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
\[e \neq 0 \rightarrow 2.71 \neq 0 \]

Answer 83

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!

Answer 84

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.

Answer 85

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.

Answer 86

\[0 =(2x-2x^3) (e^{(1-x^2)})\]

Answer 87

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

Answer 88

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)

Answer 89

totally, cant escape it hey

Answer 90

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.

Answer 91

I think that will be an adequate solution..

Answer 92

factor out the first part of the problem, and claim the 2nd term as basically irrelevant.

Answer 93

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.

Answer 94

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

Answer 95

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

Answer 96

ok no problem.. the domain lies in the x-axis