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g'(3) = -3 eh...
its a cubic at its basic
centered at the the origin
g'(3) = 0 and g'(-3) = 0
x^3 derived is 3x^2 3(3)^2 = 0 27 = 0 27 - 27 = 0
3x^2 - 27 = g' x^3 - 27x +C = function of g(x); just gotta calibrate for (3,-3)
-3 = (3)^3 -27(3) + C -3 -17 + 27(3) = C C = 61 g(x) = x^2 -27x + 61 perhaps?
g(x) = x^3 -27x + 61 that is lol
Thanks a lot. I really appreciate it.
it might be wrong; i havent checked it good wnough yet :)
C = 51 maybe...
Hmm, when I graph it it the maximum and minimum isn't (-3, 3) and (3, -3) let me try with C=51
i see it too; :) gotta figure out where I looked at it wrong
g' = 3x^2 - 27 = 0 at -3 and 3 so the derivative is good
g(x) = x^3 - 27x + C if we integrate it...
(x=-3,y=3) should fit it and (x=3,y=-3) should fit it, so we solve for C with these parameters
-3 = (3)^3 - 27(3) + C and 3 = (-3)^3 - 27(-3) + C -3 - (3)^3 + 27(3) = C OR 3 - (-3)^3 + 27(-3) = C
C = 51 or C = -51 which doesnt work out to well in teh end
g(0) would also have to equal 0
C = 0
I just graphed it again and the minimum is at (3, -3) but not the Maximum
the maximum is not at (3, -3)
g(x) = x^3 - 27x ... thats it
ack!!... that aint it either.. lol
Haha, it's ok, thanks a lot for the help though. But I got a little help from someone but have no idea how hey got it, maybe you can help elaborate this. For this problem they started from this: (x+3)(x-3) g'(x)= X^2 -9 = 1/3x^3 -9 ???? Have no idea as to how they got here for this problem.
i got some idea, but have to look at it some more
g'(x) has roots at 3 and -3 so g'(x) = x^2 - 9 if we combine them; which is good; then integrate up to x^3 --- - 9x tho 3
(1/3)(x^3) - 9x + C is the family of curves for it....
C = -15 then i think (1/3)x^3 - 9x -15 ; lets check it
Now the max is right but not the min...
I did this very quickly; \(f(x)=x^3-27x-51\). Check it!!
did that first lol
it works for one , but not the other
g(0) = 0 has to be part of te graph
so adding somethig to it aint gonna help; we gotta find the right 'slope'
M(x^3 - Bx) is what we need
Dx(Mx^3 - MBx) 3Mx^2 - MB M(3x^2 - B) = 0 when x = 3 or -3
Maybe include more powers of x?
or derive again for inflection...
6Mx = 0 when x = 0 lol ... that dint work out to well
haha, thanks, i really appreciate the help
k(-3)^3 = 3 -27k = 3 k = -1/9 maybe? x^3 ---- ; 3x^2/9 = x^2/3 i wonder 9