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anonymous
 5 years ago
Find the local extrema and inflection points of g(x)=e^(x^2/2).
anonymous
 5 years ago
Find the local extrema and inflection points of g(x)=e^(x^2/2).

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is 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 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dx = dy/du * du/dx y(u(x)) = e^((x^2)/2) where y=e^u and u=x^2/2.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dy/du = e^u du/dx = 2x/2 = x y' = x * e^((x^2)/2) thats the first derivative.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y'' the second derivative tells us y' is max or min; and also were the inflection point are: y'' = () MAX y'' = (+) MIN y'' = 0 inflection

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y'' you will need to find by the product rule: fl' + f'l

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y'' = [x][x * e^((x^2)/2)] + [1][e^((x^2)/2)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for g'(x)=xe^(x^2/2), how to i find x?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0make g'(x) equal to 0 to find your critical points...the bends of the graph.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just don;t know how to find the value of x in e^(x^2/2) :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when i equated the equation to zero

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, coz i cant think of any value that can make e to zero

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since e^x cannot ever reach 0 there is no value of "x" that will make it = 0....so it doesnt matter :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so how would the graph look like? would it be like concave up?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0concave up from (INF,1), concave down (1,1), concave up (1,INF)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the local max. is at x=0?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0There 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)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when x = 0; y'' = e^(...) y'' is () at x=0 which tells us that it is a MAX

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when is y'' = 0? for our inflections. At x=1, cause e^(...)  e(..) = 0 so, at x=1 we have a change of direction.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=x^3+y^33xy=0 (this is not a function) so do i need to use implicit differentiation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or i just differentiate it normally? it will be like =f'(x)=3x^23x=0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sure. 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you show me how you think the steps go; and I will let you kow if you are right...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=x^3+y^33xy=0 f'(x)=3x^2+03x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0explain to me: (d/dx)(y^3) = 0 cause thats an error.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes it; but your last term is in error... look at it again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes haha lemme correct it 3(y+y'x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0YES!!  now solve this equation for y'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f'(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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y'=3x^2+y'(3y^23x)3y y'y'(3y^23x)=3x^23y y'(13y^2+3x)=3x^23y y'=(3x^23y)/(13y^2+3x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as long as you kept track of yourself; you got it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so do i need to find the x values in terms of y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0equate the whole equation to zero?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0nope; 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but they did nt give me any values..: sketch the curve x^3+y^33xy=0, indicating all extrema and points of inflection. Be sure to show all of your work.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or use what we got :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0little hint; y' = 0 when that top part = 0...try that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03x^2 = 3y y=x^2 when does n^2 = n?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(0)^2 = 0 and (1)^2 = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do these make the bottom go to 0? if so, toss whichever one does.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but as is, we have bends at (0,0 and (1,1) does that make sense to you?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0to find the y'' just derive each side again with respect to x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your gonna have to use the quotient rule; and keep track of yourself tho :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats such a longgggggggggggggggggg equation...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol..... do it, and eat your brussels sprouts :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0top is: (1 3y^2 +3x)(6x 3y')(3x^2 3y)(6yy' +3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the bottom is: (1 3y^2 +3x)(1 3y^2 +3x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0im going with you work...so if its wrong... :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y"= [3y'(1+5y^2+3x+6x^2y)+3x(26y^2+9x)9y]/(13y^2+3x^2)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uhhh i guess its right coz my eyes even got dried when i was differentiating it lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it might be easier to work with the unfactored forms :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it is it is..... at (0,0) y'' = 3 we got a max at (0,0)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at (1,1) we get y''=+6 we have a MIN at (1,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got y"=3y' when (0,0)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if i plug in 0 for y', it will be zero!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0use the stuff I put up, its alot easier to work it in chunks that way

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0less degree for error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay! but why it would be max when y"=3? wouldnt it be min?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(1)(0)  (0)(3)  (1)(1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we good at (0,0) = y'' = 3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you plug in y'=0?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thin of it like this: y = (x^2) is an upside down bowl: y' = (2x) y'' = (2) = MAX

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes, I plug in (0,0) and y'=0 into what "I" did :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if i also plug in y'=0, it would be like: [(0)(1)+(0)(2)0]/ (1)(1), so wouldnt it be y"=0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i undersand the max thing now btw:) thanks

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.... 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 :)
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