What is the reverse function of    f(x)=(x^3)/3-(x^2)-1   ?
538371
http://en.wikipedia.org/wiki/Cubic_equation
294274
well, (1/3)x^3 equals x\n\nso you can simplify it to (8/3)x + 2 = f(x)\n\nthat is a good start.
121200
y = 2x is the function then the inverse is x = 2y (switch x and y) there fore F(x) ^-1 is inverse
766600
You have to use the quotient rule for this problem:\n\nFor any function f(x) defined by, f(x) = g(x)/h(x),\n\nf'(x) = [g'(x)h(x) - h'(x)g(x)]/[h^2(x)]\n\nIn this problem, we have:\n\ng(x) = xe^-x\nh(x) = x^3+x\n\nTherefore, \n\ng'(x) = xe^-x(-1) + e^-x(1) [you have to use product and chain rule for thie derivative]\n\nSimplfying, you get e^-x(1-x)\n\nh'(x) = 3x^2 + 1 (power rule)\n\nNow that we have all parts of the derivative, just plug in:\n\nf'(x) = [e^-x(1-x)(x^3+x) - (3x^2+1)(xe^-x)]/(x^3+x)^2\n\nNow just simplify this ugly mess =P\n\n---------\n\nHope this helps
1036430
In mathematics, an inverse function is in simple terms a function which "does the reverse" of a given function. More formally, if f is a function with domain X, then f −1 is its inverse function if and only if for every x &#92;in X we have:\n\n    f^{-1}(f(x))=f(f^{-1}(x))=x.&#92;,\n\nFor example, if the function x → 3x + 2 is given, then its inverse function is x → (x−2) / 3. This is usually written as:\n\n    f&#92;colon x&#92;to 3x+2\n    f^{-1}&#92;colon x&#92;to(x-2)/3\n\nThe superscript "−1" is not an exponent. Similarly, as long as we are not in trigonometry or calculus, f 2(x) means "do f twice", that is f(f(x)), not the square of f(x). For example, if : f : x → 3x + 2, then f 2 : x = 3 ((3x + 2)) + 2, or 9x + 8. However, in trigonometry, for historical reasons, sin2(x) usually does mean the square of sin(x). As such, the prefix arc is sometimes used to denote inverse trigonometric functions, e.g. arcsin x for the inverse of sin(x). In calculus, f n(x) is the nth derivative of f.\n\nIf a function f has an inverse then f is said to be invertible.
506639
The first derivitive of f(x) = 5/x^3 - 3/x^1/3 is...\n\nf'(x) = -15/x^4 + 1/x^(4/3)\n\nHope this helps. I used the product rule. Remember, you can write the term 5/x^3 and 5*x^-3. I find it simpler to rearrange the equation and use the product rule to get at the first derivitive.
696517
Y= 5x+3\ny-3=5x\n(y-3)/5= x\nso the inverse function is:\nf^{-1}(y)= (y-3)/5\n\nx
604301
We just interchange the x's and the y's. then find what is y\nExample the inverse of f(x)=x+1\nSince f(x) is just equivalent to y... \nthen y=x+1. we then switch...\nx=y+1... then find for y... y=x-1. \nThat is the inverse of f(x)=x+1\nNow let's solve your problem\ny=f(x)=7x-2\ny=7x-2\nswitch x and y \nx=7y-2\nsolve for y\n(x+2)/7=y\ninverse is (x+2)/7
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
948694
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x). \n\nIf f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1)  + C ... we need to add an arbitrary constant.\n\nIn your case we have:  f(x) = x^1  and  F(x) = (x^2)/2 + C
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
65726
Note 0: Sin inverse x is also known as arcsin(x). I'm going to use that name because it's easier to work with when you are typing answers.\n\narcsin() is the function that reverses taking the sin of a number. That is, if a = sin(b) then b = arcsin(a). In english, arcsin(y) is the angle x, that has y for it's sin.\n\nNote 1: Since  -1 <= sin(x) <= 1, arcsin() is only defined on the interval [-1, 1]. \n\nNote 2: For each value y, in [-1, 1] there are a lot of values of x such that sin(x) = y, so arcsin() is defined to be the value that is between -pi/2 and +pi/2.
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
717628
f(x) = -x^3 - x^2 + x - 9\nf(-1) means replace x with -1\nf(-1) = -(-1)^3 - (-1)^2 + (-1) - 9\nf(-1) = -(-1) - (1) - 1 - 9\nf(-1) = 1 - 1 - 1 - 9\nf(-1) = -10\n\nf(0) means replace x with 0\nf(0) = -(0)^3 - (0)^2 - 0 - 9\nf(0) = 0 - 0 - 0 - 9\nf(0) = -9\n\nf(2) means replace x with 2.\nTry this one on your own.