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

1. nincompoop

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

2. nincompoop

use latex

3. Owlcoffee

$(2x-2x^3)e ^{(1-x^2)}$

4. anonymous

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..

5. nincompoop

why would you need a calculator?

6. anonymous

sorry, whats latex ?

7. anonymous

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

8. nincompoop

you probably need to use log

9. anonymous

Yeah I was worried about that..

10. anonymous

To get rid of that E^ ?

11. Owlcoffee

No, 1 and -1 are not zeros.

12. nincompoop

they are zeroes

13. anonymous

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

14. nincompoop

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

15. Owlcoffee

He may have looked at the graphical representation.

16. 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.

17. nincompoop

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

18. Owlcoffee

Nin, don't be so rude.

19. nincompoop

@UnkleRhaukus @UsukiDoll @rational

20. 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: $2x-2x^3=0$ $e ^{1-x^2}= 0$

21. 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

22. Owlcoffee

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

23. anonymous

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

24. 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.

25. amoodarya

hint : $e^{something}\neq0$

26. Owlcoffee

Do you mean that it does no have zeroes?

27. anonymous

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

28. nincompoop

can you prove that $$e^{1-x^2} = 0$$

29. Owlcoffee

I can.

30. anonymous

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

31. Owlcoffee

oh no, wait, log 0 is not possible.

32. nincompoop

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

33. anonymous

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

34. Owlcoffee

Sorry, thought there would be a constant.

35. nincompoop

this is beyond my skill set

36. 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?

37. anonymous

lol, it cant be

38. 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: $(2x-2x^3)$

39. nincompoop

n0

40. nincompoop

I am interested to learn this

41. 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: $2x=0$ $1-x^2 =0$

42. anonymous

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

43. nincompoop

uh...

44. anonymous

lol, maybe not..

45. nincompoop

stop trolling

46. UsukiDoll

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

47. anonymous

solve for x

48. UsukiDoll

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

49. 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. $(2x-2x^3)=0$ $e ^{1-x^2}=0$

50. nincompoop

good luck

51. 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

52. 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 >:/

53. UsukiDoll

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

54. nincompoop

the easy part is always the best part, yes?

55. Owlcoffee

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

56. UsukiDoll

O____________________________________o

57. 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

58. UsukiDoll

do you mean $e^{1-x^2} = 1$ ?

59. anonymous

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

60. anonymous

or E

61. nincompoop

I still do not get E what is E

62. anonymous

the exponential constant = E

63. nincompoop

so it is not Euler's constant?

64. UsukiDoll

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

65. anonymous

sorry, yes Eulers

66. UsukiDoll

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

67. nincompoop

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

68. UsukiDoll

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

69. nincompoop

:)

70. Owlcoffee

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

71. 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

72. UsukiDoll

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

73. 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?

74. 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..

75. 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

76. UsukiDoll

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

77. UsukiDoll

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

78. 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.

79. UsukiDoll

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

80. 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?

81. anonymous

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

82. 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 $e \neq 0 \rightarrow 2.71 \neq 0$

83. 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!

84. 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.

85. 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.

86. UsukiDoll

$0 =(2x-2x^3) (e^{(1-x^2)})$

87. 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

88. 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)

89. anonymous

totally, cant escape it hey

90. 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.

91. anonymous

I think that will be an adequate solution..

92. anonymous

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

93. 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.

94. 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

95. 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

96. UsukiDoll

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

97. UsukiDoll

|dw:1434530843604:dw|

98. UsukiDoll

|dw:1434530877368:dw|

99. anonymous

So I am thinking, within the domain there is no way for E^anything to become 0

100. UsukiDoll

well e itself is 2.71 and e^0 is 1

101. 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.

102. UsukiDoll

yes the zeros are found in the other part of the equation .

103. anonymous

Thanks Usuki.. you're awesome. and thanks to everyone who came to help.. much appreciated.

104. 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 .

105. UsukiDoll

I think I did way more than the Qualified Helper... if you've paid for this question... you got ripped off :(

106. UsukiDoll

actually it's @Owlcoffee and me doing all the work

107. UsukiDoll

but still if I had to pay for qualified help, I would expect to have extraordinary result and expert explanations, not chit chat :/

108. anonymous

lol, well, you saved the day then :)

109. Owlcoffee

Sorry, i wish i could've helped better :(

110. anonymous

true, sometimes the help is not so helpful.. it costs. 10 a question I think

111. UsukiDoll

sorry 10 owlbucks is \$2 I think... you didn't pay that much did you? I notice your question was highlighted in yellow

112. UsukiDoll

@Owlcoffee you did a good job too.

113. anonymous

owlcoffee, yes you did great.

114. 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

115. anonymous

actually. that email address they say you get a refund on.. doesnt work.. it bounces.

116. UsukiDoll

:(... but that highlighted yellow question meant that you've paid for qualified help right?

117. Owlcoffee

You paid for a qualified helper!?

118. UsukiDoll

his question was highlighted in yellow... unusual since free questions are in grey

119. anonymous

yes, I just figured it was too late at night and bad timing.

120. UsukiDoll

=( get a refund...pm a moderator and explained what happened. You've been ripped off

121. anonymous

I pay for most my questions.. they're not always handled too well. but often I get a very conscientious helper..

122. UsukiDoll

not for this one... this qh was chit chatting and barely did anything.

123. UsukiDoll

right @Owlcoffee ?

124. 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.

125. 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.

126. 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.

127. 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

128. 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.

129. UsukiDoll

it depends on the question...there are some topics that I've forgot though since it has been a very long time .

130. anonymous

got about 5 weeks to get through the rest of it.

131. UsukiDoll

is this a summer school course?

132. anonymous

yeah

133. 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.

134. anonymous

its a disaster of a curriculum for calculus using mathematica, but someone has butchered it up, and everything is out of order..

135. anonymous

at least my professor is pretty cool.. having an retriceof a teacher would be awful

136. UsukiDoll

is it calculus 1,2,3, or 4?

137. anonymous

I think it';s an entry level its 234 business calc

138. UsukiDoll

o-------o. Oh I'm horrible at Business Calculus . I used to be a business major.

139. 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.

140. 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..

141. UsukiDoll

Mathematica is a computer program.. I had to use it for Calculus II

142. 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.

143. UsukiDoll

hmmm that could be the definition of the derivative... I had to use that long beast before using the real derivative techniques.

144. UsukiDoll

$\frac{f(x+h)-f(x)}{h}$

145. Owlcoffee

$\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }$

146. anonymous

yes, that part was ugly.

147. anonymous

lol, man was that ugly

148. UsukiDoll

that looks like mean value theorem xD or the rate of change formula

149. Owlcoffee

it is the definition of derivative. just a variation of it, there are 3, I think.

150. anonymous

is the other the slope formula?

151. Owlcoffee

By definition, the derivative gives you the slope of the tangent line on any point of a function.

152. anonymous

I guess thats uski's difference quotient up there too = slope

153. anonymous

okay.. time to ask another Q ...

154. UsukiDoll

I might be away from keyboard. I have to do something.

155. anonymous

starting a new thread.. all good.