How do u know if a f unction is CONTINUOUS OR NOT?
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A function is continuous if you could trace the function without ever having to lift your pencil. Every point of the function connects with every point before and after. No holes, no asymptotes, no jumps.\n\nThe formal way of saying this is that a function is continuous on an open interval (a,b) if for every c such that a<c<b, f(c) exists and the left-hand limit of f(x) as x approaches c equals the right-hand limit of f(x) as x approaches c and both limits are equal to f(c).\n\nBut my pencil analogy basically says the same thing.
764395
Hey, firstly I need this corrected: f(x)= Ix^2=5x+6I, because there should not be an equal sign after the power of 2.... I can't do this w/o knowing... if it is subtraction than I will tell you that from what I rembember about this stuff is that absolute value functions are continuous but not differentiable I think.... Because at the point, the f(c) equals the f(x) which means it is cont. but the derivative of f(x) approaching the vertex from the left side does not equal the derivative of the f(x) from the right side...whcih means it is NOT differentiable.... sorry I can't remember more buddy
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(f(2) - f(0))/(2 - 0) = 2, but f'(x) < 0 for every x, except x=1 where it is undefined.\n\nThe function has to be continuous in [0,2] and differentiable in (0,2).  For this f(x), it is undefined at x=1, so the conditions of the MVT are not met.  Math is still safe!\n\nCheers!
22349
For all real numbers, f(x) is the sum of two continuous components (as x^2 is always non-negative) and one component discontinuous at 0. Therefore the function is only discontinuous at x=0. For x>0, f(x)=x^2; for x<0, f(x)=x^2-2x.\n\nBoth x^2 and sin(x) (as well as sin^2(10x)) equal zero when x=0 and are continuous across all reals. Therefore no Squeeze Law is necessary. This expression evaluates to 0.
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GO TO AN APPRASER AND THEY CAN TELL YOU IF IT IS OR NOT
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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!
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define f(x)=d(x,A)/[d(x,A)+d(x,B)].\nThis is a continuous function f:X->[0,1] (use triangle inequality to show x->d(x,A) is continuous for a closed set A). Also f=0 on A and f=1 on B. Let C=f^(-1)[0,1/3) and D=f^(-1)(2/3,1]. Then C and D are open because inverse images of open sets under continuous functions are open.\n\nNote to Adolfo Boy: It is possible for the distance between two disjoint closed sets to be 0. Consider the x-axis and the y=1/x curve. Your argument does work for compact sets, though. The point (a,b) in your argument need not exist if the infimum is not attained.
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Homework?\n\nSuppose f is not identically 0. Then there is an interval where f stays away from 0. The g be the step function of this interval to get a contradiction.
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Given: f,g: R → R continuous\nSo,|x - c| < &#92;del ==> |f(x) - f(c)| < &#92;epi.             (i)\nand\n|x - c| < &#92;del ==> |g(x) - g(c)| < &#92;epi.                (ii)\n\nNow,\nLet h(x)=max {f(x),g(x)}.\n\nTo Prove: h(x) is continuous.\n\nProof:\nTo show h(x) to be continuous it is enough to show that the following holds...\nfor |x - c| < &#92;del ==> |h(x) - h(c)| < &#92;epi.           (iii)\n\nObserve from (i) and (ii) that \nfor |x - c| < &#92;del ==> |max{f(x),g(x)} - max{f(c),g(c)}| < &#92;epi                                                   (iv)\n\nSince f(x) and g(x) is continuous in (x - &#92;del, x + &#92;del) interval. Max{} is bounded. And hence (iv) is true.\n\nSo by (iii) h(x) is continous.
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Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
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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).
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The first one isn't, because it's undefined when x=4 (assuming the (x-4) is the denominator, not just the x)\n\nThe second one is; it's a horizontal line with no breaks or holes.
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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
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This statement can be proven by proving the contrapositive: If f:R -> R is neither strictly increasing nor strictly decreasing, then f is not a 1-1 differentiable function.\nIf f is differentiable, then f being neither strictly decreasing nor strictly increasing is equivalent to f being a constant function over some nondegenerate interval in its domain. Thus f cannot be 1-1.\nIf f is 1-1, then the condition implies that f is not continuous, and thus is not differentiable.
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Sorry I'm not english spoken\n\nIl suffit de choisir deux vecteurs u et v non nuls et non liés.\nu ∧ v ≠ 0\nOn constate alors que :\n(u ∧ u) ∧ v = 0 ∧ v = 0\nSupposons que :\nu ∧ (u ∧ v) = 0\nAlors\nu et u ∧ v sont liés\nCe qui est absurde car :\nu ∧ v ⊥ u\nDonc :\nu ∧ (u ∧ v) ≠ 0\nD'où :\n(u ∧ u) ∧ v ≠ u ∧ (u ∧ v)
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Since continuous means without stopping, the correct answer would be one.
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if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed