math: 101

Originally posted by fearlessleader
i understand (the first parts at least)

new line: how dose one find the system max of a system of non-linear inequalities?

you do this the same way you find a system max for anything non-linear--you differentiate!
then you check the boundaries for maxima that may not appear in the derivative.
not difficult.

Originally posted by BarefootChessPlayer
you do this the same way you find a system max for anything non-linear--you differentiate!
then you check the boundaries for maxima that may not appear in the derivative.
not difficult.

good.🙂
but what the hell dose that mean?😕

Originally posted by fearlessleader
good.🙂
but what the hell dose that mean?😕

you don't know how to differentiate??!
what are they teaching young people these days?!
finding derivatives is the first thing one learns in calculus.
suppose your function is y = 4*x^3 -3*x^2 + x - cos x, and the interval is [-2, 5].
first, find the derivative. the derivative of x^n is n*x^(n-1) for all real n, and the derivative of cos x is -sin x. thus, we get: y' = 12*x^2 - 6*x + 1 + sin x.
set y' equal to zero since maxima and minima of a function only occur at those points.
the polynomial has no real roots, but sin x is zero at x=0 + n*pi (here, n is an integer). unfortunately, there is no place where the derivative is zero so you have no max or min. (sometimes it works like that.)
now we check the boundaries: x=-2 and x=5. for minus two, we get -46 - cos (-2) (which is just cos 2), and for 5, it's 430 - cos 5.
by this reasoning, you only have a min at -2 and a max at 5. (if the interval had been the entire real number set, there would have been no max or min since it goes large negative for large negative x and large positive for large positive x but the cosine would cause what are called inflection points to appear infinitely often (the graph will "wiggle" at those points, integral multiples of pi).
does it make sense now?

Originally posted by BarefootChessPlayer
you don't know how to differentiate??!
what are they teaching young people these days?!
finding derivatives is the first thing one learns in calculus.
suppose your function is y = 4*x^3 -3*x^2 + x - cos x, and the interval is [-2, 5].
first, find the derivative. the derivative of x^n is n*x^(n-1) for all real n, and the derivative of ...[text shortened]... (the graph will "wiggle" at those points, integral multiples of pi).
does it make sense now?

i did not actualy read that poast, because i did not understand the fourth sentance out of about twenty.

i have no clue what they are teaching me, they cirtantly arn't doing it very fast. i'm not in calc right now, i'll be taking that next year, or maby this summer. right now i'm in trig.

what's an interval?

Originally posted by fearlessleader
i understand (the first parts at least)

new line: how dose one find the system max of a system of non-linear inequalities?

Depends very much on what the equations are. Sometimes it's easy, sometimes you'll need a computer. Sometimes there is no unique solution either:

A^2 + B^2 = 1
A + B = 1

has two solutions; B=0 and B=1

Ton

Originally posted by fearlessleader
i did not actualy read that poast, because i did not understand the fourth sentance out of about twenty.

i have no clue what they are teaching me, they cirtantly arn't doing it very fast. i'm not in calc right now, i'll be tak ...[text shortened]... maby this summer. right now i'm in trig.

what's an interval?

ok.
an interval is a range of values the variable may take on. for most operations involving real numbers, the "interval" is -infinity to +infinity.
in this case, i picked the interval [-2,5]. the brackets indicate that these "end" values are to be included in the calculation (and this is thus called a closed interval), while (-2,5) (using the bowlegs instead of brackets) denotes an open interval in which the variable does not take on the "end" value but can be arbitrarily close thereto. think of it as "-2 <= x <= 5" for the former and "-2 < x < 5" for the latter; the "equals" makes all the difference. of course, you can have one side of the range open and the other closed, as in [-2,5), depending on what you need to do.
sorry i went over you head in the previous discussion; i didn't realize you hadn't had calculus yet.

Originally posted by BarefootChessPlayer
ok.
an interval is a range of values the variable may take on. for most operations involving real numbers, the "interval" is -infinity to +infinity.
in this case, i picked the interval [-2,5]. the brackets indicate that these "end" values are to be included in the calculation (and this is thus called a closed interval), while ...[text shortened]... i went over you head in the previous discussion; i didn't realize you hadn't had calculus yet.

Hmmm....if you only understood how to derive, you would be much better off. Barefoot, although he gave an awesome explanation, went all mathy on you! (You still did better than I could do!) I will try to explain it to make more sense! A derivative, is the slope of the tangent line. Hopefully you know what slope is (y/x or rise over run, of a line). A tangent line is a line that touches a curve in only one spot. Imagine the top hill of a roller coaster: which would looks something like a sin x curve or maybe even y = -x^2. If you wanted to put a flat platform on the top of it, and only have it touch the r.c. on only one spot, it would have to have a slope of zero. That is a horizontal line. That is why, to find the min or max, you derive the equation (finding the slope of the tangent line), and then set it equal to zero, since the tangent line at a min or a max HAS TO EQUAL 0! So in the case of y = x^2, after you use barefoots instructions to derive it, you get y' = 2x. You then set that equal to zero, and solve for x. In this case, you get x = 0. You then go back to the original and plug and chug for y. The min value for y = x^2 is then (0,0), the second zero being the y-value that I got when I plugged the x back in. Hope this helps. I tried to avoid alot of the mathy language, and hopefully still kept all the logistics right. Smarter guys correct me if I am wrong here! 😀

Originally posted by !~TONY~!
Hmmm....if you only understood how to derive, you would be much better off. Barefoot, although he gave an awesome explanation, went all mathy on you! (You still did better than I could do!) I will try to explain it to make more sense! A derivative, is the slope of the tangent line. Hopefully you know what slope is (y/x or rise over run, of a line). A tangen ...[text shortened]... and hopefully still kept all the logistics right. Smarter guys correct me if I am wrong here! 😀

excellent explanation.
for some reason, the simple aspect did not come forth when i tried to do it. taht's why i can't teach.
sometimes my mind works in strange ways. *g*

Sorry for me being so formal, but a formal definition of a derivative is:

f'(x) = lim{y->x} ( f(x) - f(y) ) / (x-y)
which is the same as:
lim{dx->0} ( f(x+dx) - f(x) )/dx

so for f(x)=x^n you get
f'(x) =
lim{dx->0} ((x+dx)^n - x^n) / dx =
lim{dx->0} (x^n + n*x^(n-1)*dx + [something]*dx^2 - x^n) /dx =
lim{dx->0} n*x^(n-1) + [something]*dx =
n*x^(n-1)

this is how I learned it when I was 15.

But must say not all calculations are easy this way.

i think i was taught this in a way similar to Tony's explenation.
Example of a regularity, name it derivation, calculation rules, and then proof of product/quotient/chain rule. I do tihnk i got a little bit of what Fiathahel says, but not that formal i tihnk. Maybe just for left/right derivatives..

derivitive i get, f(X)=aX^b+cX^d
f'(X)=abX^(b-1)+cdX^(d-1)

mins and maxes are found by setting the derivative =0, poinst of inflection are found by setting f''(X)=0.

interval is the same as domain, it's all the values of X for which there is a real value for y. yes?😕

how do you find the point on the graph of a nonlinerar equation for which A is greatest if A=X^2+Y^2 or something like that?

Originally posted by fearlessleader
derivitive i get, f(X)=aX^b+cX^d
f'(X)=abX^(b-1)+cdX^(d-1)

mins and maxes are found by setting the derivative =0, poinst of inflection are found by setting f''(X)=0.

interval is the same as domain, it's all the values of X for which there is a real value for y. yes?😕

how do you find the point on the graph of a nonlinerar equation for which A is greatest if A=X^2+Y^2 or something like that?

Domain is what you set, alls values x on wich the function in question is defined (alls x for wich there exists an outcome of your function).

Interval is litterally an interval on the 'line' of numbers. An interval indicates on what part of the axis you're looking, the interval can be less then the domain, more, and also equal.

A function is defined on it's domain, but not nessesarily on an interval.

example: 1/x has domain all reals but 0, and it's not completely defined on, let's say, [0,1].

To find maxima of multi-variable equations you need to derive according to every variable on it's own, and put all those things equal to zero, that way you get a set of equations you need to solve. In your example you'd need to solve:

2X =0 and 2Y =0 ---> X=0 and Y=0, so A =0 is the minimum.