1 if x rational, 0 if x irrational?
930668
I don't feel like doing the review, but I can offer something.  It turns of that there are functions can are LeBesgue (sp?) integrable but no Riemann.  The f(x) = 0 x rational, =1 x irrational is an example.  You can cook up a set that lands on just the rationals so that it has a limit at a given rational, and hence every rational even though both the rationals and irrationals are infinitely dense in any small neighborhood of any point.  I don't think that you can do the same with irrationals.  If I'm correct then the rationals would be continuous and the irrationals not.  But I don't think that I'm correct.\n\nYou are essentially talking about what's known as point set topology, a part of an undergraduate, or graduate, real analysis course.\n\nI've found Wikipedia to be very helpful for math concepts.\n\nI just found something that said\n\nf(x) = 0, x irrational\nf(x) = x, x rational\n\nis continuous at one point, x = 0.
905755
you must include x=0.
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
902410
Irrational numbers are those which cant be expressed as terminating decimals\n\nSQRT(3) and SQRT(2) both can't be expressed as terminating decimals.\n=> they are irrational.\n\n2 * SQRT(2) = Integer * Irrational Number =  Irrational Number\n\nSQRT(2) + 1 = Irrational Number + Integer =  Irrational Number
764329
x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
1130519
Seperate the constants - you get:\n\n2x = -3\n\nDivide both sides by 2\n\n\nx = -3/2\n\ntake the cube root of -3/2\n\nIts an irrational about -1.2599
103674
2x^2 - x = 0  { by adding 1 to both sides }\nx^2 - (1/2)x = 0  { by dividing both sides by 2 }\nx (x - 1/2) = 0 { by factoring }\n\nsolves when either x = 0 or x = 1/2 . QED
1048409
It can be anything.  If you follow the sequence to (x-x) you will get an answer of 0 for the entire equation.
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
791694
The rational roots of any polynomial:\n\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) \n\nare in the form ±p/q, where p divides a(0) and q divides a(n).\nIn your case, a(0) = -1 and  a(n) = 1, so the only rational roots possible are ±1.\nSubstitute ±1 into the equation to see if either is a root. (neither is)\n\nExample:\nx^3 + 4x^2 + 2x - 4 = 0\nThe only possible rational roots are ±1, ±2, ±4 (These numbers divide a(0), and only 1 divides a(3)=1 ).  Substituting each of these into the equation (synthetic substitution works well), we find -2 the only rational root.\n\n*****************************************\nProof by contradiction.\nLet the rational number p/q be a reduced root (i.e, q  and p are relatively prime integers) to the rational polynomial equation:\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) = 0\nand such that either p does not divide a(0) or q does not divide a(n).  [Note: without loss of generality we can assume all coefficients are integers by multiplying both sides by the least common multiple of the denominators of all the a(n)'s]\n\nThen we know that \na(n)p^n/q^n + a(n-1)p^(n-1)/q^(n-1) + ... + a(1)p/q + a(0) = 0\nmultiply both sides by q^n to get:\n\na(n)p^n + a(n-1) qp(n-1) + a(n-2) q^2p^(n-2) + ... + a(1) q^(n-1)p \n+ a(0)q^n = 0\n\nor a(n)p^n + q * f(a,p,q) = 0 for some function f of a,p,q.\nand a(0)q^n + p * g(a,p,q) = 0 for some function g of a,p,q.\n\nThis means \na(n)p^n = -q * f(a,p,q)\na(0)q^n = -p * g(a,p,q)\n\nBy the first equation, q must divide a(n) * p^n.\nBut q has no factor in common with p, so it cannot have a factor in common with p^n, so all the factors of q (ie, q itself) must divide a(n), and we assumed that q does not divide a(n), so p must divide a(0) * q^n (the only possibility left)\nBut likewise with the second equation, p must divide a(0) * q^n. But p does not have any common factors with q, so it cannot have common factors with q^n, so p must divide a(0) and we assumed that p does not divide a(0), so a contradiction arises.\nTherefore no such p/q can exist, therefore the roots are irrational.\n\nQ:\nwhy shouldn't q divide a(n) or p divide a(0)?\n\nBecause otherwise if both did then the roots COULD be rational (p/q could be a root).  If thats the case you prove/disprove it by example.  In other words, its easy to show that in your example the roots are all irrational because the only POSSIBLE rational ones are 1 and -1.  So you just substitute them in to see that they're not roots. My proof just shows that IF there are rational roots, they must be p/q where p|a(0) and q|a(n).\n\nThere is no general way to determine  by inspection that a random polynomial has all irrational roots. The only way is to check the POSSIBLE rational ones by substitution.
365867
2x - (7 + x) <= x\n2x - 7 - x <= x\nx - 7 <= x\n-7 <= 0\n\nSince we're left with a statement with no variables, and the statement is true, x = all real numbers.
857286
Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
46798
if (1/x) + 2= (c/2)\n--------------------------------\n\n1/x = (c/2) - 2\n----------------------\nx = 1/((c/2) - 2)\n       (OR)\nx = ((c/2)-2)^-1\n-----------------------
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2
944543
Okay . let see \n\nPart 1 \nVertical asymptotes correspond to the zeroes of the denominator of a rational function \nx^2+2x+1= 0 \n(x + 1)(x + 1)= 0\nx = -1 \nV.A = -1 \n\nPart 2 \nhorizontal asymptotes ; limit of f(x) as x → ∞\nlim (2x - 4) / (x^2+2x+1)\nx → ∞\nlim 2x / x^2   { 1 < 2 } \nx → ∞ \n= 0\nhorizontal asymptotes is " 0 "\n\nGoodLuck
530950
i guess they r \n1. x=0\n2. x=1/54\n3. x=2\ni know i might be wrong