how do we say that a function is continuous ?
79280
a continuous function is defined in terms of limits across its entire domain.  this can be formally shown to be equivalent to this statement:  a function is continuous if it is *differentiable* across its entire domain.\n\nthis means, if a derivative exists for every point on the function where the function is defined, then it is continuous.  so, for example, the function:\n\nf(x) = { 0 when x < 1;\n         2 when x >= 1 }\n\nhas a step change at x = 1.  so the domain of this function is (-oo, oo) (that is, negative infinity to plus infinity) but it is not differentiable at x=1.  this is because the limit of the function when approaching x=1 is different when you're coming from the right or the left.\n\nbasically, continuous just means that you can graph the function without lifting your pencil off the paper.  there are no jumps or places where you have to pick up and go to another spot on the page.  another example of a discontinuous function is f(x) = 1/x, because as you approach zero from the negative side, you have to lift up your pencil when you cross zero and go to the top of the paper.\n\nthe formal definition of continuous functions is in terms of delta and epsilon, and you can read about it on this mathworld page:\nhttp://mathworld.wolfram.com/ContinuousFunction.html\n\ncheers!
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
887624
They are functions if the first thing in each pair is different from the first thing in the other ordered pairs.\n\nA) all the first things are 1 so it is not a function\nB) all the first things are different so it is a function\n\netc.
1045860
Function y = f(x) is continuous at point x=a if the following three conditions are satisfied :\n\ni.) f(a) is defined ,\nii.)lim f(x) exists (i.e., is finite) ,\nand\niii.) lim (x ->a)  f(x)  = f(a) \n\nApplying them to your case:\n(1) f(0) is defined, being a^2 and a  in two cases.\n(2)limit also exists here\n(3) f(0)=f(0-)=f(0+)\ntherefore,\na^2=a\nor a(a-1)=0\nor a= 0 or 1.
816980
1 is false. The function is continuous.\n2 is false; see above.\n3 is true.  Hence, 4 is false, and so is 5.\nYou are right about the non-differentiability at x=3.  A function cannot be called differentiable if its slope at x- does not approach its slope at x+ as the interval approaches zero.
438959
Try here (the content is very techinical).  http://scienceworld.wolfram.com/physics/ContinuityEquation.html\nhttp://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section3/continuity.htm
930668
I don't feel like doing the review, but I can offer something.  It turns of that there are functions can are LeBesgue (sp?) integrable but no Riemann.  The f(x) = 0 x rational, =1 x irrational is an example.  You can cook up a set that lands on just the rationals so that it has a limit at a given rational, and hence every rational even though both the rationals and irrationals are infinitely dense in any small neighborhood of any point.  I don't think that you can do the same with irrationals.  If I'm correct then the rationals would be continuous and the irrationals not.  But I don't think that I'm correct.\n\nYou are essentially talking about what's known as point set topology, a part of an undergraduate, or graduate, real analysis course.\n\nI've found Wikipedia to be very helpful for math concepts.\n\nI just found something that said\n\nf(x) = 0, x irrational\nf(x) = x, x rational\n\nis continuous at one point, x = 0.
589780
The definition of continuity that the preimage of any open set is open, is elegant but not directly amenable to intuition in my opinion. Maybe an easier way of putting it (logically equivalent) is the following. It's much closer to the the usually concept of continuity but applies to any topological spaces:\n\nA function f:X-->Y is continuous if and only if it preserves limit point relationships, that is: If A is any subset of X and p a limit point of A, then f(p) is a limit point, or point, of f(A).\n\nIf every open set S containing f(p) has an open preimage f^(-1)(S), then p being the limit point of some set A means that points of A are contained in f^(-1)(S) --> points of f(A) are contained in S. The converse is also easily proved so this is an equivalent definition of continuity.\n\nThe concept of a limit point p of a set A is relatively easy to grasp. It just means there's no open neighborhood of p which doesn't contain points of A - there's no finite margin separating p from points of A.\n\nIf f is continuous, for any point p we can typically construct a sequence A having p as its unique limit point, then the definition of continuity tells us that the sequence of image points under f is f(p).
565556
This is counterexample is posted on Wikipedia; the link is below.  They don't explain all of the details, so I'll do that here.\n\nConsider the function f: R^2 -> R defined by\n\nf(x,y) = x y (x^2 - y^2) / (x^2 + y^2)\n\nI'm going to show that D_x (D_y f) is different from D_y (D_x f) at (x,y) = (0,0).\n\nThis function is continuous, because it is easy to check that lim_{(x,y) -> (0,0)} f(x,y) = 0.  Its derivatives are\n\nu(x,y) = D_x f(x,y) = y (x^4 + 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\nv(x,y) = D_y f(x,y) = x (x^4 - 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\n\nAgain, these functions are continuous, because lim_{(x,y) -> (0,0)} u(x,y) = lim_{(x,y) -> (0,0)} v(x,y) = 0.  Now to compute the mixed derivatives at (x,y) = (0,0), we need to compute the following limits.\n\nD_x (D_y f) @ (0,0) = lim_{h -> 0} ( v(h,0) - v(0,0) ) / h = lim_{h -> 0} h/h = 1\nD_y (D_x f) @ (0,0) = lim_{k -> 0} ( u(0,k) - u(0,0) ) / k = lim_{k -> 0} -k/k = -1\n\nHence these partial derivatives are *not* equal at (x,y) = (0,0) as claimed.\n\n-------------\n\nBy the way, I should point out that this gives a counter example to the famous statement\n\ndiv . curl (F) = 0\n\nIndeed, take F(x,y,z) to be that vector valued-function which is f(x,y) in the z-direction.  Then\n\ndiv . curl(F) = D_x (D_y f) - D_y (D_x f)\n\nIf you choose f(x,y) as above, then this expression is not zero at (x,y) = (0,0).
352533
Generally speaking, you call "linear" a function F (or an operator whatsoever) which satisfies the following conditions:\n\n1) F(a*X)=a*F(X) for every real number a (X is the argument of the function, whatever it be: it can be a number, a vector, a complex, a function... it doesn't matter)\n\n2) F(X+Y) = F(X)+F(Y) for every X and Y which can be taken as an argument of the function or the operator.\n\nThese two conditions can be sinthesized in the following single one:\n\nCONDITION FOR LINEARITY:\nF(a*X+b*Y) = a*F(X) + b*F(Y) for every a and b reals and for every X and Y in the domain of the function.\n\n----\n\nIn the ring of the polynomials of n variables x1, x2 ... xn, it is easily verified that the only polynomials which can be defined "linear" according to the above conditions are those in which the variables appear at the first degree only, so they are in the form:\n\nP(n) = a1*x1 + a2*x2 + ... + an*xn + a(n+1)\n\nLinear equations in n variables are obtained by setting a linear polynomial of the above type equal to 0.\n\nNon-linear equations are all other possible kinds of equations: the ones obteined by setting to 0 non-linear polynomials (you call them algebraic non-linear equations), or the ones obtained by setting ot 0 expressions which involve logarithms, exponentials etc (in that case you are dealing with transcendant equations)
740329
You will have to do an epsilon-delta proof.\nNotice that\n|f(x)-f(y)|=\n|x^3-y^3 -x+y|=\n|x-y|*|x^2+xy+y^2 -1|<=\n|x-y|*[|x|^2 +|x|*|y|+|y|^2 +1]\n<=148*|x-y|
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.
694927
What are you trying to do?  You have just defined f(x).  You have not said what to do with it.
166079
Use the fact that if A is dense in X the interior of the complement of A is empty.\n\nAssume that, I,  the interior of the complement is not empty.\nPick a point in I. There has to be an interval around that point that is contained in I. That interval has a width, w.\npick n such that 1/n < w. One of the rationals k/n has to lie within the interval. Contradiction.
1050238
Here's a good statement:\n"Every continuous curve in the plane has zero area".\nIt looks like it should be true. But it is FALSE. In fact,there is a continuous curve that fills a whole square! This curve 'intersects itself' infinitely often, so we might then conjecture\n"Every arc (a one-to-one curve) has zero area".\nBut this is FALSE also! \nFinding an example of an arc with strictly positive area is not easy, but it can be done (the usual examples require the use of Cantor sets).\n\nDepending on the level, you might be able to say:\n"Every continuous function is differentiable except at finitely many points".\nThis is also FALSE. There are continuous functions that are differentiable NOWHERE.\nOr you might try:\n"Every differentiable function is monotone on some interval".\nAgain, this is FALSE.\nOr another:\n"A function with derivative zero almost everywhere is constant".\nThe 'Devil's staircase' is a counterexample.
864853
a)  f(0) = f(0-0) = f(0) - f(0) = 0.\n\nb)  Fix k in the reals, and define f(x) = kx.\n\nc)  Suppose you have f(x)=f(y) for some reals x and y.  Then f(x)-f(y) = 0, and since f preserves subtraction, f(x-y) = 0.  Since 0 is the only element which gets sent to zero, this implies that x-y = 0, and thus x = y.\n\nI have a challenge for you: find a nontrivial function f which preserves subtraction but is not injective.  There are such functions, but they are not necessarily easy to find.  Cheers!
784932
This is an excellent question.\n\nWhen dealing with "non-routine" functions, you must go back to the definition.  In the case of a piecewise function like this, you need to check that \n\nf'(x)=&#92;lim_{h &#92;to 0} (f(x+h)-f(x))/h \n\nexists from both the left and right directions.\n\nFor your problem, from the left &#92;lim_{h &#92;to 0-} (f(2+h)-f(x))/h=2 and from the right\n&#92;lim_{h &#92;to 0+} (f(2+h)-f(x))/h=5.\n\nSince the left and right limits are not equal, the function is not differentiable at x=2.  (Note that if the "+3" we omitted, it would be differentiable and some would say that we should have verified continuity first...)\n\nChallenge:\n1.  f(x)=0, x rational\n x, x irrational\n\nis continuous only at x=0.\n\n2.  f(x)=0, x rational\n x^2, x irrational\n\nis differentiable (and continuous) only at x=0.\n\nGood luck! :)