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If you have a discontinuity and you can cancel factors in the numerator and the denominator, then it is removable. If you can't cancel those factors to get rid of the discontinuity it is nonremovable. Here is an example that shows both kinds.
f(x) = (x - 2)(x + 3) /[(x - 2) (x - 4)
There is a discontinuity at x=2 but we can cancel out(x-2) from the top and bottom.
That makes it removable. However, at x=4 there also a discontinuity and there is no way to remove that one.
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What did the asymptote say to the removable discontinuity?
Don't hand that holier than thou line to me
What is function which is has irremovable discontinuity at x-2 removable discontinuity at x2 and continuous at other points?
"Removable discontinuity" means the function is not defined at that point (it has a "hole"), but by changing the function definition at that single point, defining it to be certain value, it becomes continuous. "Irremovable discontinuity" means the function makes a sudden jump at that point. There are infinitely many functions like that; for example, you can set the function to be: f(x) is undefined at x = -2 f(x) = 0 for x < 2 (except for x = -2) f(x) = 1 for x > 2
What are the limitations of regula falsi method?
Limitations of Regular falsi method: Investigate the result of applying the Regula Falsi method over an interval where there is a discontinuity. Apply the Regula Falsi method for a function using an interval where there are distinct roots. Apply the Regula Falsi method over a "large" interval.
Is the point of discontinuity considered as critical point?
Well, honey, a point of discontinuity is not the same as a critical point in calculus. A critical point is where the derivative is either zero or undefined, while a point of discontinuity is where a function is not continuous. So, in short, they may both be important in their own ways, but they're not the same thing.
Why tan x graph is continuous within its domain and is not continuous out of the domain?
we do not check if a function is continuous or not outside it's domain."first, f has to be defined at c."Tanx is not defined where cosx=0 .ie x=pi/2 , 3pi/2 etcill try to help more here.what domain means is what can you put into a function, whereas range, which i am sure you have heard of as well, just means what you can get out of a function. that being said, lets look further into the graph of tanx. when we do, we see that the graph is discontinuous at pi/2. the reason for this is because tanx is equivalent to sinx/cosx. because of this relationship, when you put pi/2 in for x in sinx/cosx, you end up with cosx=0 which makes your denominator zero, which is undefined, which makes your graph discontinuous. because of that, you cannot put pi/2 in for x in tanx, and since the domain is what you can put into an equation, pi/2 which causes a discontinuity is not included in the domain. basicly, wherever a graph is discontinuous, it wont be included in the domain because you cant put stuff in that will make your graph discontinuous
