Question

Find the local extrema and inflection points of g(x)=e^(-x^2/2).

Answer 1

is that (-x^2)/2 or is it -x^(2/2) ? its probably the first, but the first rule of mathematics is NEVER trust a mathmetician :)

Answer 2

dy/dx = dy/du * du/dx

y(u(x)) = e^((-x^2)/2) where y=e^u and u=-x^2/2.

Answer 3

dy/du = e^u

du/dx = -2x/2 = -x

y' = -x * e^((-x^2)/2) thats the first derivative.

Answer 4

y'' the second derivative tells us y' is max or min; and also were the inflection point are:

y'' = (-) MAX
y'' = (+) MIN
y'' = 0 inflection

Answer 5

it is (-x^2)/2~

Answer 6

y'' you will need to find by the product rule:

fl' + f'l

Answer 7

y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]

Answer 8

for g'(x)=-xe^(-x^2/2), how to i find x?

Answer 9

make g'(x) equal to 0 to find your critical points...the bends of the graph.

Answer 10

i just don;t know how to find the value of x in e^(-x^2/2) :(

Answer 11

when i equated the equation to zero

Answer 12

x=0 I am thinking

there is no value for e^x that will make it 0. Think about the graph of the exponent funtion, it never reaches 0

Answer 13

do you agree?

Answer 14

yes, coz i cant think of any value that can make e to zero

Answer 15

since e^x cannot ever reach 0 there is no value of "x" that will make it = 0....so it doesnt matter :)

Answer 16

so how would the graph look like? would it be like concave up?

Answer 17

so we have a point at 0 that tells us there is something going on at x=0. lets use the second derivative to get more information

Answer 18

concave up from (-INF,-1), concave down (-1,1), concave up (1,INF)?

Answer 19

and the local max. is at x=0?

Answer 20

There is a bend at 0.... but we dont really know what it is doing until we use the g''(x) to tell us more about what is happening at x=0

y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]

we can clean this up some:

y'' = x^2 * e^((-x^2)/2) - e^((-x^2)/2)

Answer 21

when x = 0; y'' = -e^(...)

y'' is (-) at x=0 which tells us that it is a MAX

Answer 22

when is y'' = 0? for our inflections. At x=1, cause e^(...) - e(..) = 0

so, at x=1 we have a change of direction.

Answer 23

so, heres what we know:

at g(0) there is a Max; and at g(1) there is an inflection. plug in those value for x in the original to get your y spots.

Does this make sense?

Answer 24

ah~yes! thanks!

Answer 25

f(x)=x^3+y^3-3xy=0 (this is not a function)
so do i need to use implicit differentiation?

Answer 26

or i just differentiate it normally? it will be like =f'(x)=3x^2-3x=0

Answer 27

sure. is it "y" is a function of "x" implicitly? cause it could go the other way around; or even be implied by an external variable like time

Answer 28

you show me how you think the steps go; and I will let you kow if you are right...

Answer 29

f(x)=x^3+y^3-3xy=0
f'(x)=3x^2+0-3x

Answer 30

explain to me:

(d/dx)(y^3) = 0 cause thats an error.

Answer 31

for this, i treated y as a constant and i didn't use implicit differentiation. should i use it? if i use it, f'(x)=3x^2+3y^2*y'-3y*y'*x

Answer 32

yes it; but your last term is in error... look at it again.

Answer 33

woops! wrong

Answer 34

yes haha lemme correct it
3(y+y'x)

Answer 35

YES!! - now solve this equation for y'

Answer 36

f'(x) = y'

So get all you y' to one side and everything else to the other side; factor out the y' and then get y' by itself

basic algebra stuff

Answer 37

y'=3x^2+y'(3y^2-3x)-3y
y'-y'(3y^2-3x)=3x^2-3y
y'(1-3y^2+3x)=3x^2-3y
y'=(3x^2-3y)/(1-3y^2+3x)

Answer 38

as long as you kept track of yourself; you got it :)

Answer 39

so do i need to find the x values in terms of y?

Answer 40

equate the whole equation to zero?

Answer 41

nope; they will give you values for x and y; or values for y' and x; or values for y' and y. you just plug in what they give you and solve for the unknown that they dont.

Answer 42

but they did nt give me any values..: sketch the curve x^3+y^3-3xy=0, indicating all extrema and points of inflection. Be sure to show all of your work.

Answer 43

you do the same thing for "with respect to time" but the thing is (d/dt) doesnt vanish, so you keep your (dx/dt) and (dy/dt).

ahh, you solve the original for y. that can be done explicitly becasue you can solve for y.

Answer 44

or use what we got :)

Answer 45

little hint; y' = 0 when that top part = 0...try that

Answer 46

x=+/- sqrty?

Answer 47

3x^2 -3y = 0 when?

Answer 48

3x^2 = 3y

y=x^2

when does n^2 = n?

Answer 49

when n=-n or +n?

Answer 50

sqrt

Answer 51

so x=+/- sqrty?

Answer 52

(0)^2 = 0 and (1)^2 = 1

Answer 53

do these make the bottom go to 0? if so, toss whichever one does.

Answer 54

but as is, we have bends at (0,0 and (1,1)

does that make sense to you?

Answer 55

yes!

Answer 56

to find the y'' just derive each side again with respect to x

Answer 57

your gonna have to use the quotient rule; and keep track of yourself tho :)

Answer 58

thats such a longgggggggggggggggggg equation...

Answer 59

lol..... do it, and eat your brussels sprouts :)

Answer 60

top is:

(1 -3y^2 +3x)(6x -3y')-(3x^2 -3y)(-6yy' +3)

Answer 61

the bottom is:

(1 -3y^2 +3x)(1 -3y^2 +3x)

Answer 62

im going with you work...so if its wrong... :)

Answer 63

y"= [-3y'(1+5y^2+3x+6x^2y)+3x(2-6y^2+9x)-9y]/(1-3y^2+3x^2)^2

Answer 64

uhhh i guess its right coz my eyes even got dried when i was differentiating it lol

Answer 65

we know y'=0 when we plug in(0,0) or (1,1) right? so plug all that in and see it we are (-),(+), or (0)

Answer 66

it might be easier to work with the unfactored forms :)

Answer 67

it is it is..... at (0,0) y'' = -3 we got a max at (0,0)

Answer 68

at (1,1) we get y''=+6 we have a MIN at (1,1)

Answer 69

i got y"=-3y' when (0,0)

Answer 70

but if i plug in 0 for y', it will be zero!

Answer 71

use the stuff I put up, its alot easier to work it in chunks that way

Answer 72

less degree for error

Answer 73

okay! but why it would be max when y"=-3? wouldnt it be min?

Answer 74

comparing to y"=+6

Answer 75

(1)(0) - (0)(3)
----------
(1)(1)

Answer 76

we good at (0,0) = y'' = -3?

Answer 77

did you plug in y'=0?

Answer 78

thin of it like this:

y = -(x^2) is an upside down bowl:

y' = -(2x)

y'' = -(2) = MAX

Answer 79

yes, I plug in (0,0) and y'=0 into what "I" did :)

Answer 80

but if i also plug in y'=0, it would be like: [(0)(1)+(0)(2)-0]/ (1)(1), so wouldnt it be y"=0?

Answer 81

i undersand the max thing now btw:) thanks

Answer 82

.... i hate to say this; but I trust my version of it better. You might have made an error when trying to squash everything together. In fact, I am sure of it :)