what is unusual about the domain of the composite f o g for f(x)=x1/2 and g(x)= -1-[x]?
491468
If by x1/2 you mean x^(1/2), and by -1-[x] you mean -1 minus the absolute value of x, then the composite function f o g is (-1-[x])^(1/2).\n\nWhat is unusual about this is that this function is not defined for any real number x.  The expression (-1-[x]) is always going to be negative, no matter what x is.  Raising this to the power of 1/2 is the same as taking the square root of it.  Since the square root of a negative number is not defined in the real number system, the domain is the empty set.
899001
You didn't write your problem clearly.  I don't know if g(x) is:\n\n1)  x² + (7/x³) - 2, or\n2)  (x² + 7) / (x³ - 2)\n\nThe quality of the answers you receive are in direct relation to the quality of the question you write!  :-)\n\nAssuming you meant (1):\n\ng '(x) = 2x - 21x^-4\n\nAssuming you meant (2):\n\ng '(x) = (-x^4 - 21x² - 4x) / (x³ - 2)²\n\nEDITED TO ADD:\n\nHi Jeff, I got your email but couldn't respond; yahoo said your email was "unconfirmed" whatever that means!  So, I am responding to you here. Hope you see this!\n\nYOU WROTE: "I was wondering if you could explain how to work through this problem"\n\nMY RESPONSE:\n\nHi Jeff, sure I am happy to explain. Basically, when you differentiate a fraction f(x) over g(x) the rule is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" -- i.e., \n\n[ (g(x)*f '(x)) - (f(x)*g '(x)) ] / (g(x))^2\n\nSo in your problem f(x) = x^2 - 7\ng(x) = x^3 - 2\n\nSo derivative = [ (x^3 - 2)*(2x) - (x^2 - 7)*3x^2 ] / (x^3 - 2)^2\n\nThen just multiply the numerator and combine like terms to get:\n\n[ -x^4 - 21x^2 - 4x ] / (x^3 - 2)^2\n\nDid that help?  Let me know if you have additional questions!\n\n.
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
1180422
The domain for a function is all the "x" values that work in the functions.\n\nAn example is "y = x"\nIn this case you can plug anything in for "x" and get a "y"\n\nAnother example is "y = 1/x"\nHere, x cannot be zero (since you cannot have 1/0)\nTherefore the domain is all values except x = 0\n\nIn your equation, (1/x^2)  - 3x  + 2, the domain is all real numbers except x = 0.  If x were 0, you'd have (1/0) - 0 + 0 and you cannot evaluate that.
894070
all the questions relate to the same basic problem. When a function p(x) changes sign in the interval [a,b] it has to cross the real line (y=0). Therefore it has a real solution there. So, first question is solved since p(1) is negative and p(2) is positive.\n\nthe second question has the same answer, only more complicated but just a change of sign when you consider the polynomial function f(x) - g(x).\n\nthird. all non-real solutions come in pairs therefore an odd degree polynomial has at least one real solution since the number of solutions equals the number of roots. (Fundamental theorem of algebra.)
785253
So are you looking for a function with a singularity on (0,1)?  That would address the question I think.\nmaybe (x-(1/2))^(1/2)\n\nTHE FIRST ANSWER IS WRONG, I THINK, BECAUSE YOUR DOMAIN IS OPEN AND DOESN'T INCLUDE 0.
935264
Actually, the answer is *not* 4+x.\n(f/g)(x) = (16-x^2)/(4-x)\nNow, factorise the top:\n(4+x)(4-x)/(4-x).\nSo *if x is not 4* then (f/g)(x) = 4+x. However, the function is not defined at x = 4.
378109
Think about what the inverse does. i.e. f(3) * f^-1(3) = 3 better be true.  Further, f(f^-1(3)) = 3. From what you are given, does f(1/5) = 3?    I can't tell. Did you type the question completely?\n\nSecond question: g(3) = g(5).  This implies it g is not a injective (1-1) function.  Can a non-injective function have an inverse?  Say g(3) = g(5) = 100 then what does g^-1(100) equal?  3 or 5?
1112829
I hope I understood the statement of the problem correctly.\nYou are asking for a set of conditions when an area will be negative.\n\n1. Let f(x)<0 then the area must lie in the positive x domain (x>0).\n2. Let f(x)>0 the area must lie in the negative x domain (x<0).
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
1048436
As a^2 is contained in F(a), clearly F(a^2) is contained in F(a). So we need to show the reverse inclusion.\n\nWe have the chain of fields F < F(a^2) <= F(a).\n\n(where I use <= and < for subset and proper subset containment) \n\nand hence [F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]. Set n = [F(a) : F] and m = [F(a^2) : F]. The preceeding equation implies that m divides n. It will suffice to that m = n. Let f(X) = X^m + c_{m-1}X^{m-1} + ... + c_1X + c_0 denote the minimal polynomial of a^2 over F. Then f(X^2) = X^{2m} + c_{m-1}X^{2(m-1)} + ... + c_1X^2 + c_0 has a as a root. This implies that the minimal polynomial of a has degree less than or equal to 2m. So we get m divides n and n <= 2m. The only ways for this to occur are when m=n or when 2m = n. But n is odd, so n does not equal 2m. Hence m = n. This completes the proof.
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
589676
Nope.  Example: "1" is a polynomial "x" is also a polynomial (these are simple polynomials).  But "1/x" is not a polynomial.\n\nPolynomials divided by polynomials are always rational functions.
466400
0 can't be in the domain.\nAlso, -1/2 cannot be in the domain because\n(-1/2)^(-1/2) = 1/sqrt(-1/2) which is not a real number\nI'm not sure exactly how to right this:\nThe domain is any real number except 0 and any number of the form (p/q), where p and q do not have any common factors and q is even and (p/q) is negative.\nIf (p/q) is negative and reduced and q is even, then you would be taking an even root of a negative number(which is never a real number).
743736
composition means like g(f(x))?\n\ndomain or result is a sub-set of domain of inner function (f),\nrange is a subset of the range of outer function (g)
974938
I believe this is how you do it:\n\n(f o g)(x) = f[g(x)] = 3(-x^2) - 4 = -3x^2 -4\n\nNow that this formula is found, sub in 2;\n\n(f o g)(2) = -3(2)^2 - 4 = -3(4) - 4 = -12 - 4 = -16
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.