The Forum > Math & Science > Math Problem Thread
That crossed my mind too. Implied multiplication with parentheses can be misleading too, though: is f(x) "f times x" or "f of x"? The result of all this is that there's no single absolutely unambiguous way to express multiplication. It seems to me that the most widely understood ways are * and parentheses, with care taken to avoid the ambiguous circumstances surrounding each. |
For all values that are not X = 2, y = 4 - x. For value of X = 2, y = 0. Basically it's a y = 4 - X graph with an exception at value X = 2. Example. f(x) = [-1 if x < 0 ] [15 if x = 0 ] [ 1 if x > 0 ] Basically this says from the interval (-∞,0) the y value is -1. At x = 0 the y value is 15. From the interval (0,∞) the y value is 1. |
Thanks, Atrophy. 'Nother question. So, the text book I'm using vaguely teaches us that any point can have a limit (except the certain specific exceptions), and that's all fine and dandy, until it asks us to fine all the limits on a graph, and there are no existing, er, hole-based limits (by hole, I mean the point where a function can't exist) on the graphs, Ought I just put "No Limits"? Or should I list the limits that don't exist and say every other point has a limit? Or can real points even have limits? My calc teacher's been gone the past days, and the sub hasn't got a clue. Related: Say you have one of the cases where a limit can't exist, such as on a vertical asymptote, and part of the problem states that that point has a value. Does that value become the limit, even though it wouldn't exist normally? Semi-Related: My calc teacher was (when he was here) teaching the class by teaching us how to use a TI-89. Problem is, I don't have a TI-89. |
On any two variable function graph there are an infinite number of points and therefore an infinite number of limits(I think, I can't think of any functions that aren't continuous on SOME interval). Remember a limit is the value that the function approaches, not it's actual value there. To answer your question I would put the intervals where the limits do exist. Graph y = sin(π/x). In the domain (-1,1) the y-values fluctuate an infinite number of times, yielding an undefined limit for any number in that domain. I'd answer this problem with "Limits exist for all numbers on intervals (-∞,1) and (1,∞) To answer your related question, No. Graph y = 1/x. The limit as x approaches 0 is ∞, that will remain the limit regardless of if there is a provision stating that at x = 0, y = 3. TI-89...eww. TI-83 Plus here, tried and true. |
The Forum > Math & Science > Math Problem Thread
