Find the domain of the function.?
952602
This is what always helps me.\n\nA function is a machine.\n\nThe domain is everything you can put into the machine so that it doesn't break.\n\nThe range is everything you can get out of the machine.\n\nIn the case of your function, there's only one possible value of x that will "break" the function, in the sense that you'd get an infinite answer.  That value is x=0, because then you'd have 9/0 + 7, and 9/0 is infinity.\n\nSo the range is {x | x is not equal to 0}.\n\nAssuming that you meant 9/x + 7, rather than 9/(x+7).  If you meant the latter, x cannot be equal to -7.
730130
Domain = x... should be all real numbers\n\nRange = y... 2 and up... x can be a negative number, but will always end up being postive when added to two, therefore the lowest number is 2
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 .
809444
ohh ya,,i know this,,\n\n1)f(x)=((x)^2)+2\n\n2)f(x)=( ( (x-3)^(-2) ) +2)\n\nbring on more of domain and range questions,,i love those problems,,
931612
f^{-1}(y) = (y-42) / 7\ns
887664
let u = x^2\ndu = 2x dx\n\nThis turns the function in to:\n1 / (u^2 + 1) - which you can work from.
820313
If f : A-->B is surjective map, then its onto but not one-one .\nso you can write domain of f as set of all values that would give values in B.\n\nYou can write as\ndom(f) = {x : f(x)=y, x is in A, y is in B}\n\nSince the function is not one-one, there can be more than one x giving the same values of y, I dont think you need more conditions in set representation.
986272
f(x) = 3(2/x + 1)\ng(x) = x^2\n\n\nHope this helps.
837800
did u mean square root of (2-x) or 1/(2-x)??\ni wil solve both the cases\n\nif it was sqrt(2-x), the solution is like this\nto find out the domain, we need to find all the values of x for which the function is defined. for a square root function, the only requisite is that the term inside the square root should be non-negative.\nso, the condition is (2-x)>=0, ie., x<=2\nso, the domain becomes (-infinity,2]\nnote that parenthsis, ie., '(' or ')' means that the value is not included in the interval, where as square brackets, ie., '[' or ']' means that the value is included.\n\nnow, if the problem is 1/(2-x), the solution is:\nfor this function to exist, the denominator should be non-zero, which means that (2-x) should not be equal to zero.\nso, x should not be equal to 2\nthe domain is given by  (-infinity,infinity) - {2}\nhere, {2} means the set containing the element 2\nthe domain can also be denoted as  R-{2}, where R is the set of all real numbers.
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
842246
y = (x + 2)^2\n\nsqrt(y) = x + 2\n\nx = sqrt(y) - 2\n\nSo, finv(x) = sqrt(x) - 2.\n\nIt is limited to the domain x>=0.
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
757602
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
995228
Domain:  All the values of x that provides a defined value of y.  In the equation above if we plug 0 or less than 0 for x, we get an undefined answer.  Thus the domain is x>0.  \n\nIntercepts:  To find intercepts we plug y equal to zero and then solve for x.  So, lnx/x = 0\n                          lnx = 0\n                             x = 1\nThus the intercept point is (x,y) = (1, 0)\n\nAsymptotes:  An asymptote is a straight or curved line which a curve will approach, but never touch.  By looking at the graph of lnx/x, we can conclude that the graph approaches y = 0 but never actually touches 0.  So, the asymptotes is y = 0.  \n\nCritical point:  In mathematics, a critical point (or critical number) is a point on the domain of a function where the derivative is equal to zero. It is also called a stationary point.  Derivative of above function is: \n                        dy/dx = [(x)(1/x)-lnx](x^2)\n                           0 = (1-lnx)/(x^2)\n                           0 = 1-lnx\n                           1 = lnx\n                            x = e\nSo, the critical point is x = e.
754001
the domain and the range of h(x):\nthe domain : x is less than or equal to 2,the range : y is greater than or equal to zero.\n\nDomain and range for the inverse of h(x) :\nD: x is grater than or equal to zero.\nR: y is less than or equal to 2.\nNotice that the range of h(x) is the same as the domain of   the inverse  of h(x).And the domain of h(x)is the same as the range of the inverse of h(x).