free tutorial on Rang of Function / Domain of function?
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The basic concept of domain and range is pretty simple.  Here are the \nfacts:\n\n  - The domain of a function is the set of all the stuff you can \n    plug into the function.X's\n\n  - The range of a function is the set of all the stuff you can \n    get out of the function.Y's
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Take a look at this site.\nhttp://www.functionx.com/visualc/index.htm
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RegMon, a free utility at http://www.sysinternals.com
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Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
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"disp" or "display" functions\n"line" is helpful too\n\nTry to make a string x='*****' y='** **' \ndisp x,y \nETC....
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domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
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start by seperating y, u then have y=3x/2+3 then plug in the domain, or the numbers u are givin\nu'll have 0, 1.5, 3, 6, 7.5
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The best way to find domain and range is to graph the function. If you graph the function that you have given as an example, you will notice that it is undefined at x=0. Since it is undefined at this point, the domain of the function is the real numbers minus 0. The range in general is slighly more difficult to find, but in this case is easy to find. This function goes to positive infinity when x approaches 0 from the left, and negative infinity when x approaches 0 from the right. The range of this function is all real numbers.
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To write any equation in "function form", just solve for y (get y by itself on one side of the =), and replace y with f(x).\n\nI'll let you do that.  :)
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Well, I suppose you want a program that computes the trace of a matrix. The trace is simply the sum of the elements in the diagonol in a matrix. I won't show it to you with OOP since there's no reason to.\n\nint trace(int matrix[][],int numrows,int numcols) {\n\nint tracetotal = 0;\n\nif (numcols != numrows) {\nstd::cout<<"Error: Can't compute trace. Not square matrix.&#92;n";\nexit(1);\n}\n\nfor (int i = 0; i < numrows; i++) {\ntracetotal+=matrix[i][i];\n}\n\nreturn tracetotal;\n}
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Domain = x... should be all real numbers\n\nRange = y... 2 and up... x can be a negative number, but will always end up being postive when added to two, therefore the lowest number is 2
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Domain: (-infinity, +infinity)\nRange: (-infinity, +infinity)
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composition means like g(f(x))?\n\ndomain or result is a sub-set of domain of inner function (f),\nrange is a subset of the range of outer function (g)
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There is no work involved. You only need to understand what domain and range mean. Domain is what can you plug in for x, range is what numbers you can get out of the function.\n\na.\nDomain: all real numbers except -9, since f(-9) will have 0 in the denominator.\nRange: all real numbers except 0, since the numerator can never be 0\n\nb.\nDomain: all real numbers\nRange: y>=-4 (y is greater than or equal to -4). You have to find the turning point using x=-b/2a=-2. f(-2)=-4\n\nc.\nDomain: all real numbers except x=-1,3, since f(-1),f(3) have 0 in the denominator.\nRange: all real numbers.
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First, evaluate these functions to understand what they mean. Here is a listing of n and f(n)\n\n0 1 2 3 4 5 6 ...\n0 2 4 6 8 10 12 ...\n\nHere is a listing of n and g(n):\n\n0 1 2 3 4 5 6 ...\n0 0 1 1 2 2 3 ...\n\nEvaluate g • f using g • f (n) = g(f(n)) to get its listing:\n\n0 1 2 3 4 5 6 ...\n0 1 2 3 4 5 6 ...\n\nwhich provides evidence for question (a). g(f(n)) = g(2n) = n, because 2n is even.\n\nThe listing of f • g is\n\n0 1 2 3 4 5 6...\n0 0 2 2 4 4 3 ...\n\nAnalyze f(g(n)) = 2 [n/2]. What happens when n is even, and when n is odd?\n\nA function, F, is injective if different values never map to the same value. Or, the listing does not have duplicates. To prove that F is not injective is easy: just find a counterexample, that is a and b such that F(a) = F(b), but a ≠ b. To prove that F is injective you must show that in general if F(x) = F(y) then x = y. \n\nLook at the listings. If a listing has duplicates then the function is not injective. A proof is just providing a counterexample. Example: f • g has duplicates; the simplest are 0 and 1: f • g(0) = f • g(1), and 0 ≠ 1.The listing of f has no duplicates. To prove that f is injective, prove that if 2x = 2y then x = y. That should be easy!\n\nA function is surjective if the listing contains every number in the domain, that is all the numbers: 0, 1, 2, 3,...\n\nThe listing of f does not contain 1. So f is not surjective. A proof is to show that there is no n such that f(n) = 1. That's easy because f(n) is even!\n\nThe listing of g appears to contain all the numbers. Does it contain 2006? Can you find n such that g(n) = 2006? Such that n/2 = 2006? Yes, n = 4012. Similarly, for any number (instead of 2006).\n\nSo, I have provided some strategies and examples to let you now make your own proofs.
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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!
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Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9