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How do I determine where a function has a removable and a jump discontinuity?

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rowkem
#1
Oct7-08, 06:38 PM
P: 53
1. The problem statement, all variables and given/known data

I am given the following function, piecewise:

f(x) = (-x+b) (x<1)
3 if x=1
(-12/(x-b))-1 (x>1, x=/b)

I am asked:

1) For what value(s) of 'b' does 'f' have a removable discontinuity at 1?
2) For what value(s) of 'b' does 'f' have a (finite) jump discontinuity at 1? Write your answer in interval notation.

2. Relevant equations

X

3. The attempt at a solution

Honestly, I have to clue where to start; especially in regards to the removable discontinuity. I tried making the functions equal each other for the jump discontinuity but, that was way out as far as I can tell.

--------------------------

If I could at least be given a starting point to go from, it would be appreciated. I'm not asking for the answers but, I'm so lost and just staring at this thing isn't helping. Thanks,

RK
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statdad
#2
Oct7-08, 07:30 PM
HW Helper
P: 1,377
If a function has a removable discontinuity at [tex] x = a [/tex], that means these things.

[tex]
\begin{align*}
\lim_{x \to a^+} f(x) & = \lim_{x \to a^-} f(x)\\
\text{ so } & \lim_{x \to a} f(x) \text{ exists}\\
f(a) & \text{ is defined} \\
f(a) & \ne \lim_{x \to a^+} f(x)
\end{align*}
[/tex]

All the limits above are finite. Basically, if there is a removable discontinuity at [tex] x = a [/tex], the function has almost everything needed to be continuous there, but the function value is "wrong". Here is an example.

[tex]
f(x) = \begin{cases}
3x+5 & \text{ if } x \ne 10\\
20 & \text{ if } x = 10
\end{cases}
[/tex]

Here the limit at 10 is equal to 35 (because both one-sided limits equal 35), but [tex] f(10) = 20 [/tex], so the function is not continuous there. We say there is a removable discontinuity at 10 because we can do this: make [tex] f [/tex] continuous merely by doing this:

[tex]
F(x) = \begin{cases}
3x + 5 & \text{ if } x \ne 10\\
35 & \text{ if } x = 10
\end{cases}
[/tex]

i.e. - we remove the discontinuity by redefining the function at the problem point, in order to make the function continuous there.

A function has a jump discontinuity at a spot if the two one-sided limits both exist but are not equal. Here is an example - the jump discontinuity is at [tex] x = 2 [/tex].

[tex]
w(x) = \begin{cases}
2x - 1 & \text{ if } x <=2\\
10x + 1 & \text{ if } x > 2
\end{cases}
[/tex]

The limit from the left is 3, the limit from the right is 21. If you were to view the graph of [tex] w [/tex] you would see a 'jump' in the two portions of the graph at [tex] x = 2 [/tex]. What you need to do is find the values of [tex] a [/tex] and [tex] b [/tex] to make your function have the types of behavior discussed here. Good luck.
rowkem
#3
Oct7-08, 08:51 PM
P: 53
If you could please relate that back to my question, it would be appreciated. Again, no answers but, I'm still a little confused and don't know where to start.

sutupidmath
#4
Oct8-08, 02:25 AM
P: 1,633
How do I determine where a function has a removable and a jump discontinuity?

Quote Quote by statdad View Post
If a function has a removable discontinuity at [tex] x = a [/tex], that means these things.

[tex]
\begin{align*}
\lim_{x \to a^+} f(x) & = \lim_{x \to a^-} f(x)\\
\text{ so } & \lim_{x \to a} f(x) \text{ exists}\\
f(a) & \text{ is defined} \\
f(a) & \ne \lim_{x \to a^+} f(x)
\end{align*}
[/tex]


luck.
It is also the case where f(a) is not defined but the overall limit as x-->a exists. The way one can remove this is by defining the function at x=a.

P.S. I am sure you know this, but this is adressed to the OP.


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