Range of an exponetnial function isthe set of all real numbers. True or false?
445844
FALSE DUDE... exponential no.'s rn't real nos themselves
19510
It is not true for all a and b. It is true if and only if a and b are positive real numbers.
865218
It's all real numbers because there's no fraction, so you don't have to worry about a number being over zero or anything, so all real numbers, your range is also all real numbers.
696843
Check MathForum.org:\n\nQuestion:\nWhat is the quantity 0^0 (zero to the zeroth power)?  Would it be \n0, since 0 times anything is 0, or would it be 1, since anything to the 0 power is 1?  Or is this undefined, too?\n\nAnswer:\nThis, too, is an indeterminate form.  Its logarithm is 0 * infinity.\n\nMore detail from the website:\nSay you wish to introduce a new element (or maybe two elements) infinity which you wish to append to the real number system.  That is not prohibited.  After all, that is how we got from natural numbers to integers (appending negative integers and zero), and from integers to rationals (appending ratios of integers), and from rationals to reals (appending limits of convergent sequences), and from reals to complexes (appending the square root of -1).  What you end up with is not the real number system, however.  Furthermore, if you wish to define the four operations + - * and / for this new system, you probably want them to be the same on real numbers, and just add on the definitions of things like infinity + r and r/infinity, for any real number r.  Some of these work fine.  It makes sense to define:\n\n        infinity + r = r + infinity = infinity\n  (-infinity) + r = r + (-infinity) = -infinity\n                infinity + infinity = infinity\n          (-infinity) + (-infinity) = -infinity\n                       infinity - r = infinity\n                    (-infinity) - r = -infinity\n                       r - infinity = -infinity\n                    r - (-infinity) = infinity\n             infinity - (-infinity) = infinity\n             (-infinity) - infinity = -infinity\n        infinity * r = r * infinity = infinity  for r > 0\n  (-infinity) * r = r * (-infinity) = -infinity  for r > 0\n        infinity * r = r * infinity = -infinity  for r < 0\n  (-infinity) * r = r * (-infinity) = infinity  for r < 0\n infinity * infinity = (-infinity) * (-infinity) = infinity\n infinity * (-infinity) = (-infinity) * infinity = -infinity\n                       infinity / r = infinity  for r > 0\n                    (-infinity) / r = -infinity  for r > 0\n                       infinity / r = -infinity  for r < 0\n                    (-infinity) / r = infinity  for r < 0\n                       r / infinity = 0\n                    r / (-infinity) = 0\n\nWhere we get into trouble is with defining the following:\n\n        infinity + (-infinity)\n     (-infinity) + infinity\n        infinity - infinity\n     (-infinity) - (-infinity)\n               0 * infinity\n        infinity * 0\n               0 * (-infinity)\n     (-infinity) * 0\n        infinity / infinity\n        infinity / (-infinity)\n     (-infinity) / infinity\n     (-infinity) / (-infinity)\n        infinity / 0 = infinity\n    (-infinity) / 0 = -infinity\n\nThese expressions are called "indeterminate forms."  These can all have a large range of different values, depending on exactly where the "infinity" parts came from.\n\nAs a result, the system you construct is not closed under addition, subtraction, multiplication, or division.\n\nOther indeterminate forms are 0^0 and 1^infinity.\n\nSo the answer, is 0^0 is undefined or indeterminate.
653782
easy infinity +1\n\nThe limit that a function is said to approach at x = a when (x) is larger than any preassigned number for all x sufficiently near a. \n\nThe key word here is preassigned.  In this case there can never be a true infinity.  You have to define the set that you are taking to infinity.  Whether it be numbers, distances, time, space, etc.  Just by defining the set, puts a limit on your answer, which makes that limit infinity.  And any comparable limit is infinity as well, unless it's exactly the same as the first.
711149
Is it lim (x approaches 0)?
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
1074093
The function has zeroes at -6 and 0.  It decreases to negative infinity as x decreases from -6. It increases to infinity as x increases from zero.  The only relative maximum is at -4.
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
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.
1035877
The natural logarithm (over the real numbers) is defined only for positive numbers, so \n\nx-4>0\n\nin other words\n\nx>4.
448269
The domain is all real numbers x, except for x=0.\n\nYou just have to remember that the exponential function and all polynomial functions are defined on the entire real line, that is there domain is all real numbers. Therefore, the only problem that occurs is with the operation of division by x. Since you cannot divide by 0, x can never be 0. All other values are fine though.\n\nIN RESPONSE to what urtimud said below: sure you can take e^0, but if you let x=0 you are taking e^(-1/0) which is undefined. Therefore, x=0 is not in the domain of this real valued function. Note that e is being raised to the ((x^2-1)/x) power, because your screen may have chopped off that last part.
885970
Domain: All real numbers\nNo vertical asymptotes\nHorizontal asymptote: lim_x->inf 1/(x+9/x) = 0\nCritical points:\ny' = [x^2+9-x(2x)]/(x^2+9)^2\n= [9-x^2]/(x^2+9)^2\nx = -3, 3
940114
I have no idea what physical domain is. If you're talking about function domains, it's all input values that give you a real output, as in all x values that give a y value.
169470
The domain is the whole set R (all real numbers) since your function is defined for every x from R.\nSimply, you can put any real number to be x and you will have  real number as an answer.
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!!
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.