How do you find the domain of a rational expression?Greatly appreciated?
957165
The domain is the set of all values the variables can have so that the value of the expression is defined.  A rational expression has a definite value as long as the denominator is not zero, since division by zero is undefined.  So any set of values for the variables is in the domain just as long as that set doesn't make the denominator of the expression equal zero.  \n\nExample :  (x^2 + 5xy + 4) / (x - 3)(y + 7) is a rational expression.  You can calculate its value for any values of x and y you plug into it just as long as the denominator (x - 3)(y + 7) doesn't equal zero.   But (x - 3)(y + 7) = 0 if x = 3 or y = -7.  So any pair (x,y) is in the domain of the original expression unless x = 3 or y = -7.  \n(1,2), (5,10), (-100,100), (4/3, -1/2) are all in the domain.  \n(2,-7), (3,41), (-8,-7), (3,-7) are not in the domain.
956017
well, to differentiate a the given funtion with respect to                (1-4x^2)^1/2, we assume (1-4x^2)^1/2=y. and the function becomes sin^-1(4y) or 1/sin4y. we have to differentiate this with respect to y.\nso, the function, which is cosec4y, on differentiation gives the result as -cosec4y*cot4y*4\nsubstituting the value of y we get the final result in terms of x as\n-4cosec((1-4x^2)^1/2)*cot((1-4x^2)^1/2)\nthe long string in the cot is (1-4x^2)^1/2 ie., y
881529
fist domain\n\nthe number we search the log must be positive so x>-2 \n\ndomain -2<x<infinity \nfor infinity log (base3), infinite is infinite. for x=-2 we tend to x+2 =0 and log (x+2 ) = -infinite  Vertical asypmtote for x=-2 \n\nthe equation can be rewritten  h(x)  = log (base 3) (x+2/7^3)
521556
Rational expressions are just fractions with variables in them. The process for adding and subtracting is the same as with fractions. First, you want to find the least common denominator, which in this problem would be 20n. Then you multiply the numerator and denominator of each term with whatever it takes to transform the denominator into the desired common denominator.\n\n1/5n=4/20n, -3/4=-15n/20n, 7/10n=14/20n\n\nNow you can combine the numerators over the common denominator:\n\n(4 - 15n +14)/20n, simplifies to (18-15n)/20n\n\nThat's all you do.
945218
[(x^2)^2]^2\n\n(x^2 -4x +4)^2\n\nmultiply      x^2  (x^2 -4x +4)^  = x^4 -4x^3 +4x^2\n\nmultiply     -4x (x^2 -4x +4)^    = -4x^3 +16x^2 -16x\n\nmultiply     4  (x^2 -4x +4)^      =  4x^2 -16x +16\n__________________________________________\n\nthe answer should be                 x^4 -8x^3 +24x^2 -32x +16\n\ni hope that helped
940347
In this case, the domain is all real numbers, because you can use any number for x. And the range is all real numbers, because you can find an x that will produce any real number y. So you write the domain and range as follows:\n\nD = {x€R} (€ is the nearest I can find to the symbol for "is an element of")\n\nR = {y€R}\n\nor you can say \n\n-∞<x<∞\n-∞<y<∞\n\nHope that helps.
188871
By expressing the variable in terms of the function, the domain of the function can be derived for real value of the variable.  This gives the range of the variable function.\n\nLet y = f(x) = x/x+7\n\nNow, solving for x in terms of y, x =7y/(1-y)\n\nNow, if we find the domain of y, y can be any real number except 1, because that would make the denominator zero, and division by zero is not allowed.  So, if R be the set of real numbers, the domain of y is R - {1}.\n\nIn other words, y can be any real number except -1.\n\nTherefore, the range of the function is R - {1}.
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2
535346
I'm not sure if this is the whole question, but I will interpret it as if it ended with a ) in place of the [exp...\n\n9(x^3)N - 12(x^4)(N^2) - 9(x^2)N
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
1079811
It *is* awfully hard to write algebra expressions on Yahoo Answers, isn't it? Oh well, we do what we can.\n\nAddressing your first question, let's write it out like a long division problem.  You'll have to imagine the horizontal bar above the larger polynomial, because I can't format it properly here:\n\nx^2 + 3x - 2  )  x^4 + x^3 - 5x^2 + 13x - 6\n\nOur first step is to look at the first term in each polynomial, and divide the smaller into the larger.  x^4 / x^2 = x^2, so the first term we write in our quotient (above the bar) is x^2.  You can write it directly above the x^4, if you wish, or above the 5x^2 term, your preference.\n\nNow, we're going to multiply that "partial quotient" by the entire divisor (x^2 + 3x - 2), which gives us x^4 + 3x^3 - 2x^2.  Write that "partial product" below the first three terms of the original dividend (x^4 + x^3 - 5x^2).  Our next step is to subtract the partial product: (x^4 + x^3 - 5x^2) - (x^4 + 3x^3 - 2x^2) = -2x^3 - 3x^2.  We then drop down the next term from the dividend, the 13x term, and repeat:\n\n-2x^3 / x^2 = -2x (add that term to our quotient)\n\n-2x * (x^2 + 3x - 2) = -2x^3 - 6x^2 + 4x\n\n(-2x^3 - 3x^2 + 13x) - (-2x^3 - 6x^2 + 4x) = 3x^2 + 9x\n\nFinally, we drop down the last term from the dividend, -6, and repeat once more.  I'm sorry that I can't write this out the way your math book can... I really hope Yahoo Answers creates some sort of way of inputting formulas, or at least monospacing text entry!\n\n3x^2 / x^2 = 3 (so we add a +3 term to our quotient)\n\n3 * (x^2 + 3x - 2) = 3x^2 + 9x - 6\n\n(3x^2 + 9x - 6) - (3x^2 + 9x - 6) = 0... no remainder!\n\nSo our final quotient is x^2 - 2x + 3, with no remainder.\n===\nNow, when doing the 2nd problem, you'll want to write the larger polynomial, the dividend, as x^4 - 3x^3 + 0x^2 + 6x - 5.  We incorporate that extra 0x^2 term so that we never find ourselves trying to subtract unlike terms.\n\nSorry there's no better way to write this out on Yahoo Answers.  Hope that helped anyway!
445923
The domain of a polynomial function is all reals.\n\nThe domain of a rational function is limited by the domains of its numerator and denominator, and the denominator cannot equal 0. \n\nSince this is a rational function of polynomial functions, the domain could only be restricted by having the denominator being equal to 0. But the range of the denominator is y >= 3. Since there is no value of of x that can make the denominator equal to 0, the domain of the entire function is all reals.
192120
Use the chain rule. Let f(y)=y^(1/2) and g(x)=x-2. Then your expression is just f(g(x)), and the chain rule states that D(f(g(x))) = Df(g(x))*Dg(x) = [(1/2)*(g(x))^(-1/2)] * [g'(x)] = [(1/2)*(x-2)^(-1/2)] * [1].\nI suggest doing several more of these questions until you no longer need to explicitly write out f and g, but can do so in your head.
891987
The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since x-intercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes.\n\nOnce you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x – a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.\n\nYou should expect to get some complicated solutions (that is, solutions containing square roots or complex numbers, or both). You will find these zeroes by applying the Quadratic Formula for the final quadratic factor.\n\nFind all the zeroes of the following polynomial: 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 \nFirst, I'll apply the Rational Roots Test--\n\nWait. Actually, the first thing I'll do is check to see if x = 1 or x = –1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1 isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so x = –1  is a root, and I can do the synthetic division with x = –1 right now.\n\n\n\nThis leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degree-four polynomial.) Now I'll apply the Rational Roots Test to get a list of values to look at:
1121129
Domain = all x-values included in the function\nRange = all y-values included in the function\n\nBoth will be written as an inequality.  Here's an example:\n\ny = square root(x-3) + 2\n\nGraph this equation on your calucluator or sketch a graph by hand by plugging in x coordinates starting with 3.  The vertex of this graph is at (3, 2).  Therefore, the domain is:\n\nx is greater than or equal to 3\n\nThe range is:\n\ny is greater than or equal to 2.\n\nIt really helps to visualize the problem when doing these problems.\n\nGood luck!
634912
I think the problem you meant is:\n\n ((2X+1)^(1/3))/(X²-10X-11).  \n\nThe domain (for real numbers) is limited by two things:  Division by zero, and the even root of a negative number.\n\nHere the denominator can be factored into (X-11)(X+10) and that will equal zero at +11 and -10.  The cube root of a negative number is not a problem.  So the domain is all real numbers not equal to +11 or -10.
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)