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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 :)
dy/dx = dy/du * du/dx y(u(x)) = e^((-x^2)/2) where y=e^u and u=-x^2/2.
dy/du = e^u du/dx = -2x/2 = -x y' = -x * e^((-x^2)/2) thats the first derivative.
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
it is (-x^2)/2~
y'' you will need to find by the product rule: fl' + f'l
y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]
for g'(x)=-xe^(-x^2/2), how to i find x?
make g'(x) equal to 0 to find your critical points...the bends of the graph.
i just don;t know how to find the value of x in e^(-x^2/2) :(
when i equated the equation to zero
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
do you agree?
yes, coz i cant think of any value that can make e to zero
since e^x cannot ever reach 0 there is no value of "x" that will make it = 0....so it doesnt matter :)
so how would the graph look like? would it be like concave up?
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
concave up from (-INF,-1), concave down (-1,1), concave up (1,INF)?
and the local max. is at x=0?
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)
when x = 0; y'' = -e^(...) y'' is (-) at x=0 which tells us that it is a MAX
when is y'' = 0? for our inflections. At x=1, cause e^(...) - e(..) = 0 so, at x=1 we have a change of direction.
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?
f(x)=x^3+y^3-3xy=0 (this is not a function) so do i need to use implicit differentiation?
or i just differentiate it normally? it will be like =f'(x)=3x^2-3x=0
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
you show me how you think the steps go; and I will let you kow if you are right...
explain to me: (d/dx)(y^3) = 0 cause thats an error.
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
yes it; but your last term is in error... look at it again.
yes haha lemme correct it 3(y+y'x)
YES!! - now solve this equation for y'
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
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)
as long as you kept track of yourself; you got it :)
so do i need to find the x values in terms of y?
equate the whole equation to zero?
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.
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.
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.
or use what we got :)
little hint; y' = 0 when that top part = 0...try that
3x^2 -3y = 0 when?
3x^2 = 3y y=x^2 when does n^2 = n?
when n=-n or +n?
so x=+/- sqrty?
(0)^2 = 0 and (1)^2 = 1
do these make the bottom go to 0? if so, toss whichever one does.
but as is, we have bends at (0,0 and (1,1) does that make sense to you?
to find the y'' just derive each side again with respect to x
your gonna have to use the quotient rule; and keep track of yourself tho :)
thats such a longgggggggggggggggggg equation...
lol..... do it, and eat your brussels sprouts :)
top is: (1 -3y^2 +3x)(6x -3y')-(3x^2 -3y)(-6yy' +3)
the bottom is: (1 -3y^2 +3x)(1 -3y^2 +3x)
im going with you work...so if its wrong... :)
uhhh i guess its right coz my eyes even got dried when i was differentiating it lol
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)
it might be easier to work with the unfactored forms :)
it is it is..... at (0,0) y'' = -3 we got a max at (0,0)
at (1,1) we get y''=+6 we have a MIN at (1,1)
i got y"=-3y' when (0,0)
but if i plug in 0 for y', it will be zero!
use the stuff I put up, its alot easier to work it in chunks that way
less degree for error
okay! but why it would be max when y"=-3? wouldnt it be min?
comparing to y"=+6
(1)(0) - (0)(3) ---------- (1)(1)
we good at (0,0) = y'' = -3?
did you plug in y'=0?
thin of it like this: y = -(x^2) is an upside down bowl: y' = -(2x) y'' = -(2) = MAX
yes, I plug in (0,0) and y'=0 into what "I" did :)
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?
i undersand the max thing now btw:) thanks
.... 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 :)