How do u find out at which values of z a function is CONTINUOUS and the values it is not?
449611
Firstly, modulus of z is continuous at ALL z, also for z=0. Cite from attached article 1: "The real absolute value function is continuous everywhere. It is differentiable everywhere except for x = 0."\n\nMay be you mean differentiable, not continuous?\nOr you are talking about continuity, but mean "sign" instead of "modulus" ? Sign(x), which is 1 for all x>0, -1 for all x<0 and 0 for x=0 is continuous at all x except 0.\n\nIn both cases, look at the definition (see attached article 2 for continuity, 3 for differentiability.\n\nBriefly, countinuity of function f(x) in point x_0 mean than any sequence, which limit is x_0, have another limit f(x_n) equal to f(x_0).\nNow you can easily check continuity of modulus and sign.\n\nDifferentiability is a bit more complex, but attached articles have enoygh information.\n\nHope it helps.
142496
There are a number of available tools in the following website:\n\nhttp://www.functionx.com/
388704
it cannot be f(x) = a for all x, unless a=0; otherwise you would be pretending that a = a+a with a non-0 !! This doesn't happen in R.
888635
Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.
437573
Functions that maps x to x\nWhere x is an element 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.
900593
y=f(x)= a*x*e^(b*(x^2)),\n y’=a*[e^(b*(x^2))+x*b*2*x*e^(b*(x^2))],\n y’=a*[1+2*b*x^2]*e^(b*(x^2))=0,\nThe value of e^(b*(x^2)) is always >0, also ‘a’ is not =0. \nthen 1+2*b*x^2=0, while x=3, b=-1/(2*3*3)=-1/18\n& 4=a*3*e^(-3*3/18), hence a=(4/3)*sqrt(e)\nis it really MAX?\ny’(0)=a>0 means increasing function, thus YES!
47235
I'm guessing you looked on wikipedia cause all those confusing terms were there.  Don't worry, I didn't understand it either :-)\n\nThe idea of the Intermediate Value Theorem is that for any interval [a,b] on any continous function f(x), there is a point somewhere in that interval where f(x) = c for any value of c such that c is between f(a) and f(b).  If you think about what a continuous function looks like and try to come up with an exception, you'll probably be able to understand intuitively why it's true.  I don't know that you can formally prove it without all the technical terms, but you should be able to prove in informally.  Basically think about what a continuous function is.  For a function to be continuous on the interval [a,b], you have to be able to trace the graph from a to b without taking your pencil off the paper.  So if there is a point c between f(a) and f(b) for which f(x) does not equal c for at least one value of x between a and b it means there's a hole somewhere.  If there's a hole, the function is not continuous.
727467
It can be a constant or a variable.  It asks what is the behavior of the function as x approaches the "a" variable or constant.  Thus,\nlim x->a [(x - a)²]/[1 + cos (π/a x)]\n\nLet me show you.  We first get the limit:\n_______________________________________\n| Since it is the indeterminate form 0/0, we can use the \n| L'Hopital's rule.\n| lim x->a [2x - 2a]/[- π/a sin (π/a x)]\n|\n| Again, it is 0/0, so use L'Hopital's Rule,\n| lim x->a 2/(-π²/a² cos (π/a x)]\n|\n| We can now substitute x\n| = 2/(-π²/a² cos π)\n| = 2/[-π²/a² (-1)]\n| = 2/(π²/a²)\n| = 2a²/π²\n|__________________________________\n\nWe now have the limit\n= 2a²/π²\n\nNow, you can interpret this limit in 2 ways.\n\nIf "a" is a constant, then the limit of that function (of only 1 variable, x) when the variable x approaches an unstated constant "a" is the limit that we have acquired (which, in this case, is a constant).\n\nNow, if "a" is a variable, then the limit of that function (of 2 variables, a and x) when the variable x approaches the value of the variable a, is a function of a single variable, which is the limit we have acquired (which, in this case, is a function).\n\n^_^
920889
if f is differentiable in x+0 then its also continue in x_0 :\n\nf(x) - f(x_0) = f'(x_0)(x-x_0)+ E(x)\n\nlim x->x_0 then left side 0 and right side 0 thus f(x) = f(x_0) for x->x_0\nthus continous in x_0
887496
The first condition for a function to be differentiable in a point x is that the function is continuous at that point.\nit's easy to prove the function is continuous at x=1\nbecause:\nLim f(x)=Lim(x^2 + 1)= 2 for x going to 1 from left (1-)\nLim f(x)=Lim(2x)=2 for x going to 1 from right (1+)\nWe have now proved that the function is continuous at x=1\n\nWe will look at f' at x=1\nf'(x)=2x ; x<=1\nf'(x)=2 ; x>1\nthe function is differentiable in a point x=a if and only if, the function is continuous and f' exist and has the same value when x goes to point x=a from left and right\n\nLim f'(x)=2 for x going to x=1 from left (1-)\nLim f'(x)=2 for x going to x=1 from right (1+)\nnow we have proven that the function is differentiable at x=1
834638
1. horizontal asymptotes : find the limit of fn as x--> infinity:\n\ndivide the fn's num and denom by highest i.e. x^2:\nr(x) = {x + (3)}/{1  - (4/x^2)}\nhence as x--> +-inf, r(x)---> +- inf\n\n2. vertical asymptote means find zeroes of denom:\nx^2-4=0\n(x+2)(x-2)=0\nhence x= 2 and x=-2 are the answers.
730983
First, define a new function g(x)=f(x)-x. f(x) has a fixed point iff ∃x: g(x)=0. By the fundamental theorem of calculus:\ng(x) = g(0) + [0, x]∫g'(a) da\n\nLet c be sup(|f'(x)|). Then f'(x)≤c<1, and so g'(a)≤c-1<0. So for x>0:\ng(x) < g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\nAnd for x<0:\ng(x) > g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\n\nHowever, [x→∞]lim g(0) + x(c-1) = -∞ (since c-1 is negative), and since g(x) is bounded above by this function for x>0, [x→∞]lim g(x) must also be -∞. Conversely, [x→-∞]lim g(0) + x(c-1) = ∞, and since g(x) is bounded below by this functon for x<0, [x→-∞]lim g(x) is ∞. But this means that g(x) is positive for some negative value of x and negative for some positive value of x, thus by the intermediate value theorem, ∃x: g(x)=0 and therefore f(x) has a fixed point. To see that it's the only one, consider that if g(x) had two zeros then by Rolle's theorem there would exist a point where g'(x) is zero, which contradicts the fact that g'(x) is strictly negative everywhere.\n\nNote that this proof remains valid if c is merely the supremum of f'(x), instead of |f'(x)|.\n\nFor the counterexample with |f'(x)|<1 (but not sup(|f'(x)|)<1), consider f(x)={1/x+x if x≥1, 2 if x<1}. This function is everywhere differentiable ([x→1-]lim f(x) = [x→1+]lim f(x) = 2, and [x→1-]lim f'(x) = [x→1+]lim f'(x) = 0), and ∀x, |f'(x)|<1, but f(x) has no fixed points.
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
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!
599545
No. Look at f(x) = |x|. It is everywhere continuous but not differentiable with respect to x at x=0.\n\nThis is a stupid question, and you're stupid for asking it.