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

[rowkem]

Homework Statement

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.

Homework Equations

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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.

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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

 

[statdad]

If a function has a removable discontinuity at $$x = a$$, that means these things.

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

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

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

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

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

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 $$x = 2$$.

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

The limit from the left is 3, the limit from the right is 21. If you were to view the graph of $$w$$ you would see a 'jump' in the two portions of the graph at $$x = 2$$. What you need to do is find the values of $$a$$ and $$b$$ to make your function have the types of behavior discussed here. Good luck.

 

[rowkem]

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]

If a function has a removable discontinuity at $$x = a$$, that means these things.

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


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.