Question

I need help explaining the graph of f(x)=x*e^(x)

Answer 1

what about it?

Answer 2

the slope at any point is e^x+x*e^x, and it goes through the point (0,0)

Answer 3

So there is a minimum at x=-1

Answer 4

well so when x>0, the graph of the function is closer to the x part of the function since the values of x*e^(x) are larger than just e^(x)?

Answer 5

(I hope I did that product rule correctly...) Well the e^x "wins" everywhere except where x is very small

Answer 6

?

Answer 7

when x>0, so say 2..2*e^(2) is bigger than jsut plain e^2) so the x part or whatevre wins

Answer 8

the curve "looks" like an exponential, not like a linear function

Answer 9

So let's split the equation up into it's two parts and look at which side dominates the final value of the function, for x=-10, -2, 2, 10...

Answer 10

well lets break it up into x>0, 0<x<1, and x>0

Answer 11

OK...

Answer 12

so first when x<0, it seems like the x affects it the most since just plain e^(-2) is a positive value but when u do -2*e^(-2) you get a negative value

Answer 13

yes, but the exponential part of the equation keeps the curve near zero while x continues to infinity

Answer 14

do you know what the limit of a function is?

Answer 15

no im just in precalc

Answer 16

do you have a graphing calculator?

Answer 17

yes a TI 84 plus

Answer 18

graph the three functions:
y=x*e^x
y=e^x
y=x

make sure you know which is which on the graph

Answer 19

A plot is attached.

Answer 20

ok

Answer 21

woah, that's kinda awesome. but what does y=x*e^x "look like" is it more like y=e^x or more like y=x?

Answer 22

The Mathematica plot statement to make the plot was:

Plot[x E^x, {x, -6, 1.6} ]

Answer 23

well are we talking about when x<0 lets do that first

Answer 24

What does the question actually ask? does it just say "describe the curve y=x*e^x"?

Answer 25

explain how each algebraic piece of the function ( so here's the the multiplication of x right?) affects portions of the graph and in her example she splits it up as when x>0, when 0<x<1 and when x>0

Answer 26

alright, those seem like reasonable sections of the curve. so when x>0 what can we say about the curve?

Answer 27

well, I notice that say we take -3..-3*e^(-3) yields a negative answers whereas e^*(-3) doesnt not so for this reason, the curve dips below into the negative y axis? or the third quadrant?

Answer 28

right, so the curve is always positive to the right of the y axis and always negative to the left of the y axis? That's a good observation. what term causes that?

Answer 29

the multiplication of the x

Answer 30

exactly, because e^x is always positive. to the left, so for x<-1 the curve is negative but always increasing, approaching zero... what causes that?

Answer 31

the e^(x)?

Answer 32

right, as e^x gets smaller and smaller, so does the curve, this is what I meant by the e^x term "dominates"

Answer 33

for that little sliver that's left, between -1<x<0, what can we say about the curve?

Answer 34

wait can u read this real fast though and tell me if it needs altering? When x<0, the graph of the function is closer to the x part of the function. This is because e^(-2) is not a negative number but multiplying that answer by -2 yields a negative answer (a positive value multiplied by a negative value always results in a negative value.) Nonetheless, as e^(x) gets smaller and smaller so does the curve.
When 0<x<1, the graph of the function is closer to the e^(x) part of the function. This is because as x approaches 1, e^(x) is larger than x*e^(x) though when x=1, the two functions intersect at the point (1, 2.7183).
When x>0, the graph of the function is closer to the x part of the function. This is because as x approaches positive infinity, x*e^(x) is much larger than e^(x) so the graph acts like the x part. In short, the curve is always positive to the right of the y-axis and always negative to the left of the y-axis (still approaching zero.)

Answer 35

these values of x overlap each other, x>0 and 0<x>1 overlap. sorry I think I misread your comment about the sections of the curve, I would split the curve into sections x<-1, -1<x<0, x>0, this covers the whole set of numbers and doesn't overlap...

Answer 36

hm are you sure in her example thats how she split it up though

Answer 37

is her example of the function y=x*e^x or is it of another function?

Answer 38

her example is e^(x)+1/x and another is e^(x) + x

Answer 39

I mean different functions can require different areas to the sectioned off in order to describe them... I would definitely use the values I described to describe this curve. It would be hard to do otherwise

Answer 40

to be* sectioned off

Answer 41

ok so for x>0 thats fine though right?

Answer 42

:/

Answer 43

no, I would say that the curve behaves like e^x because the function is always increasing at an increasing rate, unlike x, which is increasing at a steady rate

Answer 44

ok what about what i said for x<0?

Answer 45

I would again say that the curve behaves like e^x because it remains close to the x axis, but its sign is determined by the x factor out front.

Answer 46

ok so now we just have -1<x<0 right?

Answer 47

right, where the curve is dominated by the linear factor of x out front, it is negative and the slope is also negative

Answer 48

hm so the graph of the function behaves like the x part of the function? but the values for e^(x) are bigger according to my calculator..

Answer 49

yeah, that is true, but nevertheless, the curve is behaving more like the linear factor x than the exponential e^x here. exponential functions always stay on one side of the x axis or the other, unless something else is influencing them, like here a factor of x out front

Answer 50

oh okay. one sec let me fix that and then i have another question if you dont mind

Answer 51

ok

Answer 52

what about f(x)=x^(1/x)

Answer 53

oof, is the question to describe it everywhere? or si there a specific range over which you're supposed to describe it?

Answer 54

same thing x>0, 0<x<1 and x>0 unless we need to alter it slightly like the last one

Answer 55

are you sure the function is x^(1/x)?

Answer 56

yep

Answer 57

have you been studying imaginary numbers?

Answer 58

i know how to work with them a bit

Answer 59

i know like -1 = i or something?

Answer 60

hold on, I need to get my bearing on this one a little, it's been a while...

Answer 61

ok

Answer 62

alright, so at zero this beast is undefined because you have an exponent of 1/0 which is infinity...
at x=1 the function is equal to 1, and at x=-1 the function is equal to -1...

Answer 63

ok so can we do x<0, 0<x<1 and x>0 then?

Answer 64

yeah, i'd start a new post asking someone about this, it's a complicated one... sorry not to be of more help man

Answer 65

no its okay do you have a couple minutes for a different question? sorry to hog ur time!

Answer 66

sure...

Answer 67

okay so I am given these 12 points and i have to find the equation i have the graph its basically i think a sin graph but with changing amplitude and period..are you good with any of that type of stuff

Answer 68

is this a Fourier transform (more than one sine wave) or is it just a basic Asinwx+h)+b kinda thing?

Answer 69

let me try to find a pic of the basic idea hold on

Answer 70

hmm having trouble finding a pic should i just give you the points and maybe you can get an idea from that?

Answer 71

sure

Answer 72

ok (-.75, 0), (-.5, -.7906), (-.25, -.8274), (0.0), ( .25, .90641), (.5, .94868), (1, -1.039), (1.25, -1.088), (1.5,0), (1.75, 1.1915), (2,1.2471), (2.25,0)

Answer 73

and you're sure this is a periodic solution, not a polynomial or anything else?

Answer 74

it appears as a periodic though like i said the period and amplitude change neither are constant

Answer 75

found a pic! one sec

Answer 76

http://www.ltcconline.net/greenl/courses/204/appsHigherOrder/forcedVibrations.htm

Answer 77

the graph right before forced vibrations with damping

Answer 78

alright, well, we know the period from the zero crossings right? so the period must be 0.75...

Answer 79

no isnt it 1.5?

Answer 80

oh, right, that's what I meant, thought 1.5 wrote 0.75

Answer 81

yeah so 2pi/b = 1.5 and i can solve for b right?

Answer 82

yup

Answer 83

now what though this is where im stuck my teacher thought try finding you see how there are maximum values?

Answer 84

well, the trick here is that they don't give you the maximums, but you know where they are on the x axis, they're between the zero crossings

Answer 85

ok so whats my next step since i now know "b"

Answer 86

:/

Answer 87

yeah, I'm not going to be a lot of help here, the thought I had was to regress a line through the two points on either side of a maximum, and where that intersected the x axis would give you the origin of the function, like some plus or minus factor, but I'd post this as a separate question also.

Answer 88

sorry man :/

Answer 89

its ok

Answer 90

i think this is due tomorrow though so gahh need to finish

Answer 91

could you help with j(x)=x^(2)*e^(-x)? :/

Answer 92

sorry, I need to get back to work. :/

Answer 93

okkkk