math hmwk... restrict the domain of f so that f^-1(x) is also a function?
828415
You restrict the domain of the function as necessary in order that it pass both the vertical and horizontal line test.  Once you do that, f^-1 exists.  For example, if f(x) = sinx, you can restrict its domain to (-pi/2, pi/2). This function will then pass both h & v line test, and its inverse will then exist.\n\nanother example - f(x) = x^2.  A parabola opening upward whose vertex is the origin, and fails the h-line test.  If you restrict domain to (>=0),  f^-1 will then exist.
1159041
If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings.  If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones.  If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.
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)
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.
983754
piecewise functions are differentiable only in the respective domains,not through out.otherwise they will not be piecewisewill they?
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
1031874
There are a few requirements you need to fulfill to make f(x) differentiable everywhere:\n1)  The function needs to be continuous at x=2\n2)  The limit of f(2)-  [the limit of f(x) as you approach x=2 from the left] must equal the limit of f(2)+ [the limit of f(x) as you approach x=2 from the right]\n3)  f'(x) at x=2 must equal the same number -- for both x^2 and mx+b.\n\nSo, let's solve it out.\n\n1)  To be continuous, 2^2 = 2m + b.  So 4 = 2m+b.  So b = 4-2m.\n2)  We'll check this last.\n3)  When f(x) = x^2, f'(x) = 2x.  When f(x) = mx+b, f'(x)=m.  So 2x=m.  So let 2*2 = m.  So m = 4.\n\nGoing back to (1), we have b = 4 - 2(4) ==> b=-4.\n\nLet's check (2) now.  The limit of f(2) as you approach x=2 from the left is x^2, or 4.  The limit of f(2) as you approach x=2 from the right is 4(2)-4, or 4.  The 2 limits are equal, so we're ok.\n\nANSWER:  m=4, b=-4.
1169084
Domain lists all possible x values. A normal polynomial function has a domain of all real numbers, because EVERY x value works.\n\nYour task is to find x values that DON'T work...and since you cannot divide by zero, then any x in the denominator that makes x^4+1 equal to zero will not be part of the domain.\n\nSo, what do you know about x^4+1?\n\nIt is an even power, so even if you put a negative number in for x, x^4+1 will be positive (since adding one will only make it more positive). If x=0, that is still positive, because of the plus one.\n\nx^4+1 is ALWAYS positive for any x...therefore it is NEVER equal to zero. Therefore any x value works, and your answer is all real numbers!!
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
188871
By expressing the variable in terms of the function, the domain of the function can be derived for real value of the variable.  This gives the range of the variable function.\n\nLet y = f(x) = x/x+7\n\nNow, solving for x in terms of y, x =7y/(1-y)\n\nNow, if we find the domain of y, y can be any real number except 1, because that would make the denominator zero, and division by zero is not allowed.  So, if R be the set of real numbers, the domain of y is R - {1}.\n\nIn other words, y can be any real number except -1.\n\nTherefore, the range of the function is R - {1}.
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.
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
239640
Domain of a function f:X->Y is the set of all members of X. A well-defined function must map every element of the domain to an element of its codomain. In the given function, square root(x) is defined for all x if y can be a complex number. So,  if y belongs to the set of complex numbers, the domain is the set of complex numbers. And if, y is any real number, then x is the set of al positive real numbers. For more info, read http://en.wikipedia.org/wiki/Domain_%28mathematics%29 .
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.
316755
What you have written doesn't make much sense. First, by mod z, do you mean |z|? I'm guessing you do. Then, do you mean ->C (as in a function going from {z: |z|<1} into C) or do you mean  to exclude a point C from your set?\n\nIn any case, you are being asked to show a result that is false. There are many analytic functions that won't have |f(z)| go to infinity as |z| goes to 1. Take f(z)=z, for example.
1121129
Domain = all x-values included in the function\nRange = all y-values included in the function\n\nBoth will be written as an inequality.  Here's an example:\n\ny = square root(x-3) + 2\n\nGraph this equation on your calucluator or sketch a graph by hand by plugging in x coordinates starting with 3.  The vertex of this graph is at (3, 2).  Therefore, the domain is:\n\nx is greater than or equal to 3\n\nThe range is:\n\ny is greater than or equal to 2.\n\nIt really helps to visualize the problem when doing these problems.\n\nGood luck!
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2