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

[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

 

[jbunniii]

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

 

[houssamxd]

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

 

[jbunniii]

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]

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?

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

 

[jbunniii]

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?

 

[houssamxd]

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

 

[HallsofIvy]

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).<br /> <br /> For what values of x is at least one of those NOT true? (Look specifically at (1)!)

 

[houssamxd]

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).<br /> <br /> For what values of x is at least one of those NOT true? (Look specifically at (1)!)

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

 

[jbunniii]

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]

The first example in your link is exactly like this one.

http://www.dummies.com/how-to/content/how-to-determine-whether-a-function-is-discontinuo.html

 

[houssamxd]

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??

 

[jbunniii]

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?

 

[houssamxd]

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

 

[jbunniii]

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 $x=4$.

 

[houssamxd]

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

 

[jbunniii]

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

Absolutely, good luck!