let the function f be defined by f(x)=2x-1 if 1/2f(root of t)=4 what is the value of t?
435571
Given: f(x) = 2x - 1 and 1/2*f(√(t)) = 4. \n\nThen f(√(t)) = 8 where f(√(t)) = 2*√(t) - 1. \n\nSo, we must solve the equation 2*√(t) - 1 = 8 as follows\n\n2*√(t) - 1 = 8\n2*√(t) = 9\n√(t) = 9/2\nt = 81/4.
715887
f(x) = x^5 +2x^3 -x^2 -2x -6\nf(x) has one change of sign so it has one positive root.\n\nf(-x) = -x^5 -2x^3 -x^2 +2x -6\n\nso it has got 0 or 2 -ve roots.\n\nThis is as per Descate's rule
794327
A pretty trivial one would be f(x) = absolute value of x and g(x) = -x\n\nf(g(x)) = abs (-x) = x\n\nbut g(f(x)) = - (abs x) = -x
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
700967
Restating your question (for syntax purposes):\n\nF(x) = sqrt(x-1) / e^x \n\nThe critical values are defined to be when F'(x) = 0 or when F'(x) is undefined.  Let's solve for F'(x).\n\nTo solve for F'(x) we need to use the quotient rule.\n\nF'(x) = [1/(2sqrt(x-1))*e^x - sqrt(x-1)*e^x]/(e^x)^2\nF'(x) = (e^x - 2*(x-1)*e^x)/[(e^2x)*(2*sqrt(x-1))]\nF'(x) = 1/2 * [1 - 2(x-1)]/(e^x * sqrt(x-1))\nF'(x) = 1/2 * [2 - 2x]/(e^x * sqrt(x-1))\n\nWe then calculate when F'(x) = 0 or when F'(x) is undefined.\n0 = 1/2 * [2 - 2x]/(e^x * sqrt(x-1))\n0 = [2 - 2x]/(e^x * sqrt(x-1))\n\nSo \n2 - 2x = 0\ne^x * sqrt(x-1) = 0\n\n2x = 2\nsqrt(x-1) = 0\n\nx = 1\nx-1 = 0\n\nWhich yields x = 1 as a critical value.
554760
I do not know about F.But I know (x).Does she have something like(3,4)(3,5)?\n\nx_|y_ <-- or does she have that?\n3, 4\n\ny=2f(x)\n4=2f(3)                          \n\n      ^\n      That's what she is supposed to do.\n      To solve the function.\n     But still i am sorry about not knowing F.\n    (Those # is example)
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)
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
840688
Not sure what you were doing. I would rewrite f(x) in this form:\n\nf(x) = (2x-1)^(-1/5)\n\nThen you're dealing with a u^n situation, so\n\nf'(x) = (-1/5)(2x-1)^(-6/5)*(2)\n\nf'(x) = (-2/5)(2x-1)^(-6/5) \n\nThat's equivalent to what you had, but a little simpler.\n\nLet the underboss know that this is a "site" and I did answer your question.
965701
f(x) = x(x+2)(x-2)(x-4)\n\nMultiply it out if you need it in general form.\n\nThere could also be a constant in front of the x.
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
121200
y = 2x is the function then the inverse is x = 2y (switch x and y) there fore F(x) ^-1 is inverse
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.
852324
We have f(x) = (x)^(7x) = e^(7x*ln(x)). So, f'(x) = e^(7x*ln(x)) * d/dx(7x*ln(x) = . (x)^(7x) *7 *(x *1/x + ln(x) = 7* (1+lnx) * (x)^(7x)
1051317
f(x): 'f' is usually a defined function.\n\nf(x) = |4|, would mean the the function returns the absolute value of 'x', which in this case is four.\n\nei: f(-4) = 4. If function is defined as above. written out would be in some programing languages would be f(abs(x)).\n\nf(x) can be any formula you want to define it as.\n\nf(x*y) or f(sin(x))....\n\n\nreally unsure of the how you have stated the problem.
839617
i think you meant F(0)=0\nand f(t)=6sec^2(t)-9t^3\nso\nF(t)=6tan(t)-(9/4) t^4+c\nF(0)=c=0\nF(t)=6tan(t)-(9/4)t^4