jbunniii said:Is the function even defined at ##x = \pm 4##?
Discontinuous doesn't mean the same thing as undefined. Go back to the definition of continuity: can a function be continuous at a point if it is not defined at that point?houssamxd said:well that's what u have to find
where is it discontinuous
i have an exam on this but its not clear yet to me
jbunniii said:Discontinuous doesn't mean the same thing as undefined. Go back to the definition of continuity: can a function be continuous at a point if it is not defined at that point?
OK, so you said the function is discontinuous at ##x = 4##. Referring to your link, what kind of discontinuity does it have at that point?houssamxd said:well basically we just have to use the function and find the points where its disconitinuous
i.e in the graph there is a hole or a jump
check here
http://www.dummies.com/how-to/content/how-to-determine-whether-a-function-is-discontinuo.html
0jbunniii said:OK, so you said the function is discontinuous at ##x = 4##. Referring to your link, what kind of discontinuity does it have at that point?
For ##x = -4##, what is the value of the denominator at that point?
HallsofIvy said:Are you saying that the difficulty is that you do not know what "continuous" means? Didn't It occur to you to look up the definition?
A function, f, is continuous at x= a if and only if:
(1) f(a) exists
(2) [itex]\lim_{x\to a} f(x)[/itex] exists
(3) [itex]\lim_{x\to a} f(x)= f(a).
For what values of x is at least one of those NOT true? (Look specifically at (1)!)
Note thathoussamxd said:0
but if u factorize the denominator and cancel you will only get -4 as a point of discontinuity
try it yourself
jbunniii said:Note that
$$\frac{4-x}{16-x^2}$$
and
$$\frac{1}{4+x}$$
are not the same function. They are equal for ##x \neq 4##, but the first function is undefined (hence discontinuous) at ##x=4## whereas the second is defined and continuous at ##x=4##.
Therefore the first function has what kind of discontinuity at ##x=4##?
That's right. After you remove it by canceling the common factor in the numerator and denominator, the result (second function I listed above) is continuous at that point.houssamxd said:i guess it has a removable one right??
jbunniii said:That's right. After you remove it by canceling the common factor in the numerator and denominator, the result (second function I listed above) is continuous at that point.
Now what about ##x = -4##? You correctly determined that it is discontinuous there. What kind of discontinuity is it?
Sure. A removable discontinuity is still a discontinuity. As your link describes it, you can think of it as a "gap" in the graph of the function at ##x = 4##. You can fill the gap by removing the common factor so the denominator will be defined at ##x=4##.houssamxd said:non removable
but do they both count as discontinuities
jbunniii said:Sure. A removable discontinuity is still a discontinuity. As your link describes it, you can think of it as a "gap" in the graph of the function at ##x = 4##. You can fill the gap by removing the common factor so the denominator will be defined at ##x=4##.
Absolutely, good luck!houssamxd said:thanx a lot man
i owe you one
wish me luck for tomorrows exam
A removable discontinuity is a point on a graph where there is a hole or gap in the line. It occurs when a function has a value that is undefined or not defined at a particular point, but can be filled in by redefining the function at that point.
A removable discontinuity can be identified as a point where the graph has a break or a gap, and there is a hole in the line. It can also be identified as a point where the function is undefined or not defined, but can be made continuous by redefining the function at that point.
A removable discontinuity is caused by a factor in the function that cancels out, resulting in a hole or gap in the graph. In the case of f(x)=(4-x)/(16-x^2), the factor (16-x^2) cancels out, leaving a hole in the graph at x=4.
The value of the removable discontinuity can be determined by finding the limit of the function as it approaches the point of discontinuity. In the case of f(x)=(4-x)/(16-x^2), the limit as x approaches 4 is 1/4. Therefore, the value of the removable discontinuity is 1/4.
Yes, a removable discontinuity can be removed by redefining the function at the point of discontinuity. In the case of f(x)=(4-x)/(16-x^2), the function can be redefined as f(x)=1/(x+4), which removes the discontinuity at x=4 and makes the function continuous at that point.