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.

<br /> \begin{align*}<br /> \lim_{x \to a^+} f(x) &amp; = \lim_{x \to a^-} f(x)\\<br /> \text{ so } &amp; \lim_{x \to a} f(x) \text{ exists}\\<br /> f(a) &amp; \text{ is defined} \\<br /> f(a) &amp; \ne \lim_{x \to a^+} f(x)<br /> \end{align*}<br />

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.

<br /> f(x) = \begin{cases}<br /> 3x+5 &amp; \text{ if } x \ne 10\\<br /> 20 &amp; \text{ if } x = 10 <br /> \end{cases}<br />

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:

<br /> F(x) = \begin{cases}<br /> 3x + 5 &amp; \text{ if } x \ne 10\\<br /> 35 &amp; \text{ if } x = 10 <br /> \end{cases}<br />

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.

<br /> w(x) = \begin{cases}<br /> 2x - 1 &amp; \text{ if } x &lt;=2\\<br /> 10x + 1 &amp; \text{ if } x &gt; 2<br /> \end{cases}<br />

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.

<br /> \begin{align*}<br /> \lim_{x \to a^+} f(x) &amp; = \lim_{x \to a^-} f(x)\\<br /> \text{ so } &amp; \lim_{x \to a} f(x) \text{ exists}\\<br /> f(a) &amp; \text{ is defined} \\<br /> f(a) &amp; \ne \lim_{x \to a^+} f(x)<br /> \end{align*}<br />


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.