Register to reply 
How do I determine where a function has a removable and a jump discontinuity? 
Share this thread: 
#1
Oct708, 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/(xb))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 


#2
Oct708, 07:30 PM

HW Helper
P: 1,361

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 onesided 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 onesided 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. 


#3
Oct708, 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.



#4
Oct808, 02:25 AM

P: 1,635

How do I determine where a function has a removable and a jump discontinuity?
P.S. I am sure you know this, but this is adressed to the OP. 


Register to reply 
Related Discussions  
Determine a function that calculates the nth weight factor of a wave function  Advanced Physics Homework  0  
Determine the function so that the integral holds  Calculus & Beyond Homework  6  
How to determine when a function changes sign?  General Math  1  
Removable discontinuity  Calculus & Beyond Homework  6  
How to determine cost function (I really need help on this ASAP!)  Introductory Physics Homework  1 