find global maxima for f(x)=sinx.sin2x (+infinity,-infinity)?
925132
TimGNO is correct, although a pedant would point out that he hasn't shown this point is a maximum.  But  it is common sense.  See my response to the second time you asked the question, I got the same answer (foolishly mucking around with tangents), but make a couple of other observations you may wish to look at.
146506
well, you don't have to worry about sinx and cosx as they are always between -1 and +1. so the first part of the function (1/2)* sinx is bounded.\ncos(x) is bounded.\nSo, you can rewrite the function as,\ny(x) = f(x) - x*g(x).\nwhere both f(x) and g(x) are bounded.\nso, as x goes to infinity, y(x) tends to minus infinity because of -x in the right hand side. \nSimilarly when x goes to minus infinity, y(x) goes to plus infinity\n----------------------\nResponse to the counter-example - \nin case y(x) = 2+x*(-3),\nas x -> infinity y -> -infinity.\nso y(x) is unbounded\nwhat's wrong with that??????
953356
Let f(x) = sin (1/x) and g(x)=-sin (1/x). Neither f nor g have limits at 0, but [x→0]lim f(x) + g(x) is obviously 0.
878725
antiderivative of:\n20-12cos2pi(x-0.1)\nis:\n\n20x - 12 sin( 2pi(x-0.1)) / 2pi +C\n=20x - 6 sin( 2pi(x-0.1)) / pi +C\nto double check just take the derivative of the function and see that you get:  20-12cos2pi(x-0.1) '
433978
Simply equate the functions as it asks you to;\nso you get 1-sinx = 2sin2x\nThen use the trig identity sin2x=2sinxcosx\nso you get 1-sinx=4sinxcosx\n\nCan you solve it from there?\nI'd next use cosx = sqrt (1-(sinx)^2)\nThen square the equation and solve it for x as you've probably done a million times before.\n\nYou'll probably get a quadratic equation in sinx. Use the quadratic formula to get some values out, and remember that you have to give answers between -360 and 360 degrees.
722288
(x/2)*(100-x^2)^(1/2)+\n50*arcsin(x/10)\n\nSorry it's on two lines, yahoo cuts it off if I leave it on one. arcsin is the inverse sine function.
793658
min of f(t) = 0 when t = 0 and t = 8\nmax occurs at t=2\n\nI assume you mean the maximum and minimum of the absolute value of f(t). If that's the case then the minimum of f(t) occurs when t= 0 or 8 where f(t) =0\n\nTo determine the max we find when the derivative f ' (t) = 0 \n\nlet g(t) = cuberoot(t) = t^(1/3) and h(t) = (8-t)\n\nThe derivative of f(t) (i.e. f ' (t)) is given by\n\nf ' (t) = g'h+h'g  (the product rule)\n\nf ' (t) = (8-t)(1/3)t^(-2/3) - t^(1/3)\n\nThe max, min and inflexion points occur when f ' (t) =0\n\nt^(1/3) = (8/3)t^(-2/3) - (1/3)t^(1/3)\n\n(4/3)t^(1/3) = (8/3) t^(-2/3)\n \nt =2\n\nf(2) = 7.56 apprx is the max
746686
well, here goes nothing.  i have an answer, but i'm not 100% sure if it's right.  as you said, (V0/g) is a constant.  the derivitive of sin(2x) is 2cos(2x).  set equal to 0 and i got:\n(V0/g)2cos(2x)=0\ndivide both sides of equation by g/2V0:\ncos(2x)=0\nthen i graphed the function of cos(x) and saw that it equalled 0 at (pi/2) and (3pi/2).  we're looking for minimum so we keep (pi/2) and discard (3pi/2).  since we're dealing with 2x and not x we have:\n2x=(pi/2)\ndivide both sides by 2\nx=(pi/4) or 45 degrees.  hope this matches up to the back of the book
804449
There is no limit. The problem is that sin(x) goes between postive one and negative one infinitely many times as x gets small (i.e. goes to negative infinity). The hard part, though, is to figure out how to show this.\n\nI will look at two different sequences of numbers that go to negative infinity. When I plug the numbers in the first sequence into x/(2sin(x)) they will go to positive infinity. When I plug the numbers in the second sequence into x/(ssin(x)) they will go to negative infinity. If I can get two different answers then there is no limit.\n\nThe first sequence is x_n = -pi/2, -pi/2-2pi, -pi/2-4pi, -pi/2-6pi, ....,-pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-pi/2-2n pi)/(2sin(-pi/2-2n pi))=(-pi/2-2n pi)/(-2)=pi/2+2n pi\nAs n gets big then this number goes to positive infinity.\n\nThe second sequence is x_n = -3pi/2, -3pi/2-2pi, -3pi/2-4pi, -3pi/2-6pi, ....,-3pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-3pi/2-2n pi)/(2*sin(-3pi/2-2n pi))=(-3pi/2-2n pi)/(2)=-3pi/2-2n pi\nAs n gets big then this number goes to negative infinity.\n\nWe now have two sequences of numbers that give different answers. The function cannot converge to a single number.
986056
the above is maximum when ax^2+bx+c is minimum\n\nax^2+bx+c is minium or c when x = 0 as a,b are positive\n\nso maximum value f(x) is at 0 that is A exp(-c)
1183491
Lim { [ sin(x-1)  ] / [ x^2 + x - 2] }\nx-->1\n\nAdd the following to the denominator\n-2x + 2x +1 -1\n\nLim { [ sin(x-1)  ] / [ x^2 + x - 2 -2x +2x +1 -1] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ x^2 -2x + 1 +3x -3] }\nx-->1\n\nNote: (x^2-2x+1) =(x-1)^2    and  (3x-3) = 3(x-1)\n\nThen\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)[ (x-1)+3 ] }\nx-->1\n\nLim { [ sin(x-1) / (x-1) ] * [ 1/ (x-1+3 )] }\nx-->1\n\nLet u= x-1 then as x-->1, u--> 0\nand\nLim { [ sin u / u ] * [ 1/ (u+3 )] }\nu-->0\n\nSince\n\nLim  [ sin(u) / (u) ]=1\nu-->0\n\nThen \nLim { [ sin u / u ] * [ 1/ (u+3 )] } = 1 * [1/(0+3)] = 1*1/3 = 1/3\nu-->0
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
870571
F(x)=2[sec(x)^2] * [tan(x)] / [xcos(x)+2sin(x)]\n\nremember:\nsec(x)=1/cos(x); tan(x)=[sin(x)/cos(x)]\n\nWe rewrite the function and get:\nF(x) = [2/cos(x)^2] * [sin(x)/cos(x)] / [xcos(x)+2sin(x)] <=>\n\nF(x) = [2 /cos(x)^2] * [sin(x) /{ cos(x) * [xcos(x)+2sin(x)]} <=>\n\nF(x)= 2sin(x) / [ x*cos(x)^4 + 2sin(x)cos(x) ] =>\n\nLimF(x)= 2sin(x)/[x+2sin(x)] \nx→0\nBecause all terms with cos(x)→1 IF x→0 =>\n\nLimF(x)=2sin(x)/[x+2sin(x)] = 2sin(x)/x \nx→0\nbecause 2sin(x)→0 IF x→0 =>\n\nLim F(x)=2 because sin(x)/x →1 IF x→0 \nx→0\n=============================\nRegard from Denmark
909799
gopal's solution:\n\nf(x)=x^3-6x^2+9x+2\nf'(x)=3x^2-12x+9=0 for extremas\n3(x^2-4x+3)=0\n3(x-3)(x-1)=0\nx=3 or 1\nf''(x)=6x-12\nat x=1 6(1)-12=-ve so max\nat x=3 6(3)-12=+ve so min\nso the absolute maximum is at x=1 and\n=6\nabsolute min is at x=3 and is=2  <------ However still need to check endpoints of the interval [-1,4] as these turnng points are actually LOCAL maxima and minima!\n\nSo f(-1) = -1 - 6 - 9 + 2 = -14 < 2\nSo ABSOLUTE minimum = -14 at x = -1\n\nAlso f(4) = 64 - 96 + 36 + 2\n= 6\nSo ABSOLUTE maximum = 6 and occurs at both x = 1 AND at x = 4
943798
x + y = 20, so y = 20 - x\nyou want to maximize/minimize y + √x = 20 - x + √x\n\nf(x) = 20 - x + √x\ntake the derivitive and set = 0.\n\nf '(x) = -1 +1/(2√x)\n-1 +1/(2√x) = 0\n1/(2√x) = 1\n2√x = 1\n√x = 1/2\nx = 1/4 = .25\n[Here, y = 20 - x = 19.75.  19.75 + √(1/4) = 19.75 + 0.5 = 20.25 is the maximum value of the sum y + √x ].\n\nThis is a maximum.  The second derivitive, f ''(x) = (-1/4)x^(-3/4), is negative for all x>0 and thus the function is concave downward everywhere in that interval.\n\nThe minimum will therefore be at x= 20, the end of the interval.\n[Here, y = 0.  y + √x = 0 + √20 ≈ 4.472 is the minimum of the sum]
919994
Find the 2nd derivative and find the max.
697951
find out what happens to Y (what it approaches) as:\nx goes to infinity\nx goes to negative infinity\nthen find out what happens to X as:\ny goes to infinity\ny goes to negative infinity\n\nalso find the intercepts (if they exist)