beginning composite functions?
956480
Composition of functions means that you evaluate one function, then take that answer and use it to evaluate the next function.  For example, if you are trying to find (f o g)(1), plug 1 into the formula for g(x).  Then take that answer and plug it into f(x).  If you are trying to find (g o f)(-2), plug -2 into the formula for f(x) and then use that answer in g(x).  To find (g o g)(-1), plug -1 into the formula for g(x), then use that answer in g(x) again.\n\nsome websites for math:\nwww.algebrahelp.com\nwww.askdrmath.com\nwww.purplemath.com
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
991053
Substitute or plug in the value for x an solve the problem using PEMDAS\n\nParentheses\nExponents\nMultiplication&\nDivision\nAddition&\nSubtraction
875755
Try tracing a very small example on paper to gain some insight into this.  Perhaps just two or three elements with at least two equal keys.  Then reverse the order of the two equal keys in the input set, and see if you still get the same result.  Tracing the algorithm like this will get you the answer in a manner that should give you some confidence.
177983
simple chk out:\n\nhttp://www.cplusplus.com/doc/tutorial/structures.html
668107
First derivatives are usually represented like this\nf_x = ( f(x+h,y) - f(x,y) )/ h\nf_y = ( f(x,y+h) - f(x,y) )/ h\n\nYour second partials should be:\n\nf_xx = ( f_x(x+h,y) - f_x(x,y)) / h\nf_yy = ( f_y(x,y+h) - f_y(x,y)) / h\n\nYour mixed partials are:\n\nf_xy = ( f_x(x,y+h) - f_x(x,y)) / h\nf_yx = ( f_y(x+h,y) - f_x(x,y)) / h\n\n\nThat is f_x is a function of x and y. To get f_xy take that functions derivative with respect to y.
387163
Use "..."\n\nI don't really get it, but there's some documentation here: http://courses.cs.vt.edu/~cs1704/notes/A02.AdvancedFunctionParameters.pdf
271377
Now, if only they would enable the user to input their OWN functions...that would be useful.  \n\nBTW, if anyone knows how to do that, I would surely like to know!
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
238393
I will Answer On A Sample :\n\n    Delegate Function fns_Operation(ByVal par1 As Double, ByVal par2 As Double) As Double\n\n    Function fns_01(ByVal par1 As Double, ByVal par2 As Double) As Double\n        Return par1 + par2\n    End Function\n\n    Function fns_02(ByVal par1 As Double, ByVal par2 As Double) As Double\n        Return par1 * par2\n    End Function\n\n    Function fns_Operate(ByVal par1 As Double, ByVal par2 As Double, ByVal fns As fns_Operation) As Double\n        Return fns(par1, par2)\n    End Function\n\n    Sub Test()\n        MsgBox(fns_Operate(123, 343, AddressOf fns_01))\n        MsgBox(fns_Operate(3453, 24234, AddressOf fns_02))\n    End Sub
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
751599
Primarily this is an expression raised to the fifth power, and so that is the outer function:\nf(x) = x^5,  or f(u) = u^5\n\nClearly, to get your function, you  need to replace x by\nx^2 + 4x + 6, and so that is the inner function, g(x).\nChecking, note that\nf(g(x)) = f(x^2 + 4x + 6)\n           = (x^2 + 4x + 6)^5\n\nI expect you know the "chain rule" for differentiating a composite function:\ndy/dx = (dy/du)*(du/dx)\n         = 5u^4*(2x + 4) and you finish this by replacing\nu with x^2 + 4x + 6 and taking out the common factor from \n2x + 4.
120442
Because of the way you would call the function \nsomefunct(param1, param2, param3); // canonical\nsomefunct(param1, param2); // compiler knows you mean to use the default param for 3 because it´s the third param you don´t provide.
635332
It sounds like a function that was expecting two or more parameters, was only passed one.\n\nfunctionx(x,y)\n\nwas only passed 1 parameter:\n\nfunctionx(x)\n\nFind the function and give it what it needs!\n\nGood luck.
152511
Which function, Subsitution, Linear COmbination, Graph?
524475
It can be done using Taylor series (see first attached article)\n\nFunction f(x) can be approximated near some point x_0 by a sum of such series:\nf(x_0) + \n+ (x-x_0) * f'(x_0)/1! +\n+ (x-x_0)^2 * f''(x_0)/2! + \n+ (x-x_0)^3 * f'''(x_0)/3! + ...\n\nHowever, as you can see from section "List of Taylor series of some common functions" of first article, and section "Newton's generalized binomial theorem" of second article, binomial formula is correct not only for positive, but also for negative or even for complex p!\n\n\nThe only thing you need to do is generalize the concept of bnomial coefficients.\nUsually, they are counted as:\nC(n,m) = n!/(m!(n-m)!), but factorial are defined only for positive integers.\n\nUsing an alternative defiinition it becomes possible to operate with any real n:\nC(n,m) = n * (n-1) * (n-2) *...* (n-k+1) / k!.\nEasy to see, that for positive integer n those formulas are equivalent.
984829
basically, think of a function as a machine:\n\ntake this an an example: f(x) = 2x +3\nsee the x in brackets?\n\nif x = 1, then f(1) = 2(1) + 3 = 5\nif x = 2, then f(2) = 2(2) + 3 = 7\nif x = 3, then f(3) = 2(3) + 3 = 9\n\nsee whats happening? the answer DEPENDS on number inputted in the brackets.\n\nso therefore, the definition you quoted "a consistent, predictable, relationship between the independant and dependant variable" is true.