what is the minimum value of X + 1/X=?
471125
Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
374365
it will be choice (c)\n(x+1)(x-1)\nx^2+x-x-1\nx^2-1
197379
x is anything less than 1.\n4 x1=4 so 1 cannot be the answer (unless the < had a line underneath it)\n4x0=0 which is less than 4\n4x -2= -8
757845
x  y\n0 0\n1 -1/5\n-5 1
832985
Let S = x + 1/x.\n\nWe want to minimize x for x>0.\n\nTake the derivative of S with respect to x and set that equal to zero to find the extrema of S:\n\n0 = dS/dx = 1 - 1/x^2\n\n1/x^2 = 1\n\n1 = x^2\nx = 1 (and -1, but we are only interested in x>0 here).\n\nWe can check to make sure this is a minimum by looking at the sign of the second derivative of S with respect to x at x=1:\n\nd2S/dx2 = 2/x^3\n\nWhen x = 1, the second derivative is equal to 2, which is positive, so the extremum at this point is a minimum (the function is concave upward).
951594
Let x be one number\nother number=100+x\nproduct\n=x[100+x]\n=100x+x^2\ndifferentiating\nf'x=100+2x\nthis is a minimum \nwe put this=0\n100+2x=0\nx=-50\nthe numbers are\n50,-50
101082
(x-1)/4 > 3\nx-1 > 12\nx > 13
581319
obviously 1 is right. Now read the solution:\nf(x)=x+1/x (x > 0)\nf'(x)=0 then 1-1/x^2 =0 so x=+/-1\nsince x>0 then the only root is x=1\nbut we must check if this could be min point (and not a Max point). We can check it by evaluating numbers greater and less than 1 and see how the result grows. But mathematically, we have a function f'(x)=1-1/x^2, which is a semi-elliptic curve, having two roots at +/-1 and the slope of function is negative at x=1 (since f"(x)=-2/x^3 equals -2 at x=1). Therefore, the x=1 is the minimum point.
932401
the answer is 0\n\nu know, lim 1/(n+1) = 0\n              n -> ∞ \n\nand, lim (n/n+1)^n = 1\n        n -> ∞ \n\nso: lim (1/n+1)(n/n+1)^n = lim (1/n+1) * lim(n/n+1)^n = 0 * 1 = 0\n      n -> ∞                           n -> ∞          n -> ∞
822539
first multiple everythingby x, then simplify and solve:\n\n1/x-2 = 3/x+3 =>\n\nx/x - 2x = 3x/x +3x =>\n\n1 - 2x = 3 + 3x =>\n\n-2  = x
984623
It will be 1/(x^2)\n\nFor example\n\n(1/2)^2 = (1^2)/(2^2) = 1/ 4\n(1/3)^2 = (1^2)/(3^2) = 1/ 9\n(1/4)^2 = (1^2)/(4^2) = 1/16\n\nSo the result actually gets smaller even though you are squaring it
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
892981
since this limit is in a indeterminant form of 0/0 you can apply L'hospital's (sp?) rule\nwhich is taking the derivitive of the top and the derivitive of the bottom and evaluating the limit\n\nso this becomes [1/(1+x)]/1\nwhich is Lim x->0 1/(1+x) \n= 1/1 = 1\n\nedit:\nbah in that case im not sure...\nmaybe try a table of values and show that from each side of zero the limit approaches 1
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
914435
x > 5\n\n- - - - - -s-
1122465
Just plug in 1 for x.\n√(1-1) = √0 = 0