# Removable Discontinuity of f(x)=(4-x)/(16-x^2)

• houssamxd
In summary, the given function, f(x) = (4-x)/(16-x^2), has a removable discontinuity at x = 4 and a non-removable discontinuity at x = -4. This means that the function has a "gap" or "hole" in its graph at x = 4, but the gap can be filled by removing the common factor in the numerator and denominator. The function is not defined at x = -4 and cannot be made continuous at that point.
houssamxd
removable discontinuity

## Homework Statement

the following function
f(x)=(4-x)/(16-x^2) is discontinuous at?

i got at -4 but some of my friends say its 4, -4

how is that possible

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Is the function even defined at ##x = \pm 4##?

jbunniii said:
Is the function even defined at ##x = \pm 4##?

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

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
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 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
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?

jbunniii 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?
0

but if u factorize the denominator and cancel you will only get -4 as a point of discontinuity

try it yourself

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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) $\lim_{x\to a} f(x)$ exists
(3) $\lim_{x\to a} f(x)= f(a). For what values of x is at least one of those NOT true? (Look specifically at (1)!) 1 person HallsofIvy said: Are you saying that the difficulty is that you do not know what "continuous" means? Didn't It occur to you to look up the definition? A function, f, is continuous at x= a if and only if: (1) f(a) exists (2) [itex]\lim_{x\to a} f(x)$ 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)!)

but if u factorise it and cancel the common
you only get -4 as the point of discontinuity

houssamxd said:
0

but if u factorize the denominator and cancel you will only get -4 as a point of discontinuity

try it yourself
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##?

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##?

i guess it has a removable one right??

houssamxd said:
i guess it has a removable one right??
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?

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?

non removable
but do they both count as discontinuities

houssamxd said:
non removable
but do they both count as discontinuities
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##.

1 person
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##.

thanx a lot man
i owe you one
wish me luck for tomorrows exam

houssamxd said:
thanx a lot man
i owe you one
wish me luck for tomorrows exam
Absolutely, good luck!

1 person

## What is a removable discontinuity?

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.

## How can you identify a removable discontinuity on a graph?

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.

## What causes a removable discontinuity?

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.

## How can you determine the value of the removable discontinuity?

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.

## Can a removable discontinuity be removed?

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.

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