f(x) = x^2 / (x - 2 ) ^2?
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
887664
let u = x^2\ndu = 2x dx\n\nThis turns the function in to:\n1 / (u^2 + 1) - which you can work from.
819082
Let f(x) = x ³/ (x² + 2x - 15) \n\n= x - 2 + (19x -30)/(x² + 2x - 15) (by division)\n\n= x - 2 + (19x -30)/[(x - 3)(x + 5)]\n\nSo f(x) is undefined when x = -5, 3\n\nAlso lim x→±∞ f(x) = x - 2\n\nThus f(x) is asymptotic to g(x) = x - 2 as x →±∞ and asymptotic to x = -5 and x = 3
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
567979
inverse of f(x) is (x-1)/2\n\nworking:\nlet inverse of f(x) be y\n       2x+1  = y\n       2x    = y-1\n        x    = (y-1)/2\nthus, inverse of f(x)= (x-1)/2
771554
(x+5)(x-5)\n(x)^2-(5)^2\n=x^2-25
902917
y=1/x^2-3 => y+3 = 1/x^2 => x= +/-sqrt(1/(y+3))
860831
I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)
794664
1) If h(x) = f(g(x)) = (x^3 -1)/(x^3 +1), then \nf(x) = (x-1)/(x+1) and g(x) = x^3\n2) If h(x) = f(g(x)) = 3sqrt(x^2 -8), then\nf(x) = 3sqrt(x) and g(x) = x^2 -8
681267
Don't think so.\n\n(d^2/dx^2)(1/x^2) = d/dx(-2/x) = -2lnx\nbut\n(d/dx(1/x^2))^2 = (-2/x)^2 = 4/x^2\n\nSo not necessarily equal.
600301
x^3+x^2+2 = x^5+2\n\n(x^2+2)^2 = x^4+4\n\n(x^3+x^2+2)/[(x^2+2)^2] = (x^5+2)/(x^4+4) = (x+2)/4 = (x+1)/2
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
1147786
x - sqrt(x) = 0\nx = sqrt(x)\nx^2 = x\nx(x-1) = 0\nx = 0,1\n\nBoth check.
832910
yes the integral of the derivative of f(x)=x^2 is x^2.\n\nand just to check:\nf(x)=x^2\nf'(x)=2x\nF(x)=2*(1/2)x^(1+1)=x^2\n\nwhether you graph it or not makes no difference.
1054506
given, F(x)= x-[x]\n=> F(1/x)= 1/x- [1/x]\n\nthen equation to be solved is :\n\nx-[x]+ 1/x -  [1/x]=1\n\nor x + 1/x = 1 + [x] + [1/x]\n\nnow when x>0, [x] = x \nthen from above equation,\nx+ 1/x= x+1/x+1  this is not possible for any positive x.\n\nwhen x<0,  [x]= -x\n\nthen,\nx+ 1/x = -x -1/x +1\nor, x+ 1/x = 1/2, this has no solution such that x is negative.\n\nHENCE the above equation has no solution.
800627
x = 3/(x-2)\n\n=> x(x-2) = 3\n=> x^2 - 2x = 3\n=> x^2 - 2x - 3 = 0\n=> (x-3) (x+1) = 0\n\nEither x-3 = 0 => x=3\nOr x+1 = 0 => x=-1
110508
(x^2+5) / (x^3+15-2x^2+1)^2\non [0,1]