coupla calc questions, answers to any would be a great help?
927535
#1 \n\nIf f(x)= √(x³ + 5x +121) * (x² +x +11) \n\nthen, f(0)= √(121) * (11) \n= 11 * 11\n= 121\n\nf'(0) = -1/2 (See #3 below at ***)\n\n#2 \nIf y is a differentiable function of x, then the slope of the curve xy^2 -2y + 4y^3 = 6 at the point where y = 1 is what?\n\nWhen y = 1\n\nx(1)^2 - 2(1) + 4(1)^3 = 6 \nie x + 2 = 6\nSo x = 4\nSo y^2 + 2xydy/dx -2dy/dx + 12y^2dy/dx = 0\nSo 1 + 12 dy/dx - 2dy/dx + 12 dy/dx = 0\nThus 22dy/dx = -1\ndy/dx = -1/22\nie slope of tangent at (1, 6) is -1/22\n\n#3\nFind all open intervals on which the following function is decreasing:\nf(x) = x / (x^2 + x -2)\n= x/[(x + 2)(x - 1)]\n= 2/[3(x + 2)] + 1/[3(x - 1)] (by partial fractions)\nDiscontinuities at x = -2 and x = 1\n\nf'(x) = -2/[3(x + 2)²] - 1/[3(x - 1)²]\n\nSo f'(0) = -2/[3(2)²] - 1/[3(-1)²]\n= -2/12 - 1/3\n= -1/2 ***\n\nClearly f'(x) < 0 for all x except x = -2 and x = 1 where it is undefined \n\nie the function is decreasing fo x< -2, -2 < x < 1 and x > 1
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html
889431
1. Find the antiderivative of 2x + 2sin(x).\nIt is f'(x) = x^2 - 2cos(x) + C\nSince f'(0) = 2, then you can plug in 0 for x and set this equal to 2.\n0^2 - 2cos(0) + C = 2\n-2 + C = 2\nC = 4\nf'(x) = x^2 - 2cosx + 4\nNow find this antiderivative.\nf(x) = (1/3)x^3 - 2sinx + 4x + C\nSince f(0) = 2, we have\n(1/3)0^3 - 2sin(0) + 4(0) + C = 2\nC = 2\nSo f(x) = (1/3)x^3 - 2sin(x) + 4x + 2\nTo find f(2) just replace x with 2 and evaluate.\n2.  Find the antiderivative of 7x +4\nIt is f(x) = (7/2)x^2 + 4x + C\nSince it passes through the point (9,5) that means x = 9 and f(x) = 5\n5 = (7/2)9^2 + 4(9) + C\n5 = (7/2)81 + 36 + C\n5 = 567/2 + 36 + C\n5 = 639/2 + C\nC = 5 - 639/2 = -629/2\nThus, f(x) = (7/2)x^2 + 4x - 629/2\nPlug in x = 3 and solve for f(x).
443281
1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
123634
http://members.aol.com/oelaineo/formulas.html
851464
that is a really simple question, so if you have a calculator you should be able to get it, otherwise, it is your homework and I refuse to give you an answer to your homework. I would actually HELP you to get the answer if you asked for HELP doing it.\n\n-EA
1098983
I will not use the chain rule for this one, because you'll just make things complicated.  Just use quotient rule:\nu = f(x)\nv = g(x)\nd (u/v) = (v du - u dv)/v² \n\nlet\nu = x\nv = sqrt(7 - 3x)\n\nd/dx (x/sqrt(7 - 3x))\n\nuse the rule\n= [sqrt(7 - 3x)(1) - x(1/2)(1/sqrt(7 - 3x))(-3)]/ [sqrt(7 - 3x)]²\n\nsimplify\n= [sqrt(7 - 3x) + 3x/2 sqrt(7 - 3x)]/(7 - 3x)\n\nsimplify\n= [2(7 - 3x) + 3x]/2 sqrt(7 - 3x)(7 - 3x)\n\nsimplify\n= (14 - 6x + 3x)/2(7 - 3x)^3/2\n= (14 - 3x)/2(7 - 3x)^3/2\n\n^_^
725168
Use L'Hospitals rule\nd(sin ax)/d( bx)\na*cos(ax)/b lim x----> 0  = a/b
864388
some formulas that can help :\nhttp://www.eskridgeinc.com/support/formulas.html
873308
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
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
343985
need to use the chain rule i.e.\n\nf'(pq) = p'*q + p*q'\n\nwe get:\nae^ax/(sin(bx+c)) - be^ax*cos(bx+c)/[(sin(bx+c))^2]\n\nThis is the correct answer\n\n*edit* srry couldnt see all of it if its on one line so split over two lines the answer is:\n\nae^ax/(sin(bx+c)) \n-be^ax*cos(bx+c) / [(sin(bx+c))^2]\n\n\nand ofcourse if you like, further simplification is possible with cos(bx+c)/sin(bx+c) =1/tan(bx+c)=cot(bx+c), also you can take a factor of e^ax out if you like but the above answer is perfectly accebtable.
713733
Problem 1: in 2005, t=2005-1998=7\nin 2008, t=2008-1998=10\namt property appreciates by between 2005 and 2008\n= f(10) -f(7)\n=75,000(1.06^10) - 75,000(1.06^7)\nfind the answer by typing the above into a calculator\n\nI don't really understand problem 2. What are you trying to find?\n\nI'm thinking it could be either P or r:\nTo find P: A=Pe^rt\ne^rt=A/P\nIn(e^rt)=In(A/P)\nrt = In(A/P)\nr = [In(A/P)]/t\n\nTo find P:\nP = A / [e^rt]\n\nI hope these were helpful.
888760
I hope you appreciate this since it is a bit of work. Hopefully it will help you solve similar problems. There are problems with the data you presented but you will get the idea of how to solve these problems.\n\n1) the cost function C(x) is:\n \nC(x) = (-x^2+165x+136)/120\n\nYou need to determine a,b,c from 3 independent equations of the form ax^2+bx+c = specified value.You can choose any 3 of the 5 data points to determine a,b,c. I chose the first 3 and determined a,b,c. However, though the above function works for x =1,4,5 it does not work for x=7 or 10. So the data is inconsistent. Also, this is not a realistic cost function since if x>165.82 the cost becomes negative.\n\nIf x =1....  a+b+c=2.5\nIf x=4...16a+4b+c=6.5\nTaking the difference between these equations gives:\n15a+3b=4 \nb = (4-15a)/3 = (4/3)-5a\n\nIf x= 5....25a+5b+c = 7.8\nor using the value of b above\n25a+5(4/3 -5a)+c =7.8\n25a+20/3-25a+c = 7.8\nc=7.8-20/3 = 39/5 - 20/3 = (117-100)/15 = 17/15 = 136/120\n\na+b+c=2.5\nSo a+4/3-5a+17/15 = 2.5\n-4a = 2.5-17/15-4/3 = (37.5 -17-20)/15 = 0.5/15\na= -0.5/60 = -1/120\n\nb=(4/3)-5a = 4/3-5(-1/120) = 4/3+5/120 = 165/120\n\n2) The average cost per unit x is:\n\n C(x)/x = -x/120 + 165/120 + 136/120x\n\nd/dx C(x)/x = -1/120 - 136/120x^2 \n\nWhen the average cost is a minimum this must = 0\n\nSo 0=1+136/x^2 but this would make x^2 = -136 which is impossible. It would still not make since if it was the Total cost which you were trying to minimize, snce, as mentioned, for large enough x this becomes negative and increasingly so.\n\n3) The total cost is given by the the function C(x). The elasicity E is the sensitivity of the total cost to a change in production.\n\nSo E = d C(x)/dx = C' (x) = (-2x+165)/120\n\nthe average cost per unit x is:\n\n C(x)/x = -x/120 + 165/120 + 136/120x\n\nThe elasticity is again given by the derivative of this funcion:\n\n= d/dx C(x)/x = -1/120 - 136/120x^2\n= (-1/120)(1+136/x^2)\n\nWe see that this elasticity is negative.
25743
in Excel, use NormDist() function to answer a and b,\nNormInv() for c and d, and read your textbook for e and f.\n\nyou will fail final exam if you do not understand how these things work, and if you want me to help you more, it will be $25/hour.
740329
You will have to do an epsilon-delta proof.\nNotice that\n|f(x)-f(y)|=\n|x^3-y^3 -x+y|=\n|x-y|*|x^2+xy+y^2 -1|<=\n|x-y|*[|x|^2 +|x|*|y|+|y|^2 +1]\n<=148*|x-y|
618077
go to www.curnaturecalculus.com  They have all the answers.\n\nSplooch!