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Math Help - Continuity; finding a value of "k"

  1. #1
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    Continuity; finding a value of "k"

    Find k so that the following function is continuous on any interval.






    My calc teacher has this habit of not explaining how to do weird problems like this. If I see it done once, I'll be able to do it with any other problem. I'm just having trouble understanding where to even start. Even "hints" would be fine...I'd like to actually *understand* this so I can do well on the tests!
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  2. #2
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    um . . . I'm not sure if this explanation will help.


    so what's the \lim_{x \rightarrow 0} cos(x)

    and what's \lim_{x \rightarrow 0} 6e^x

    So you know that the first one is 1, and the second one is 6.


    So you want to make it so they're equal since your relationship with k in the first limit is multiplication and the relationship with the second limit is subtraction:

    You want:

    k*1 = 6-k
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by lysserloo View Post
    Find k so that the following function is continuous on any interval.






    My calc teacher has this habit of not explaining how to do weird problems like this. If I see it done once, I'll be able to do it with any other problem. I'm just having trouble understanding where to even start. Even "hints" would be fine...I'd like to actually *understand* this so I can do well on the tests!
    Since both of these functions are continuous on their own, the whole function j(x) will be continuous if the two parts are equal at x=0.

    Spoiler:
    As it is now k\cos(0)=k and 6e^0-k=6-k. So to make this whole thing continuous, all we have to do is set k equal to 6-k and solve for k. So k=3 will make j(x) continuous.
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  4. #4
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    Okay so let me ask you a question, redsoxfan325, if that's okay.

    I understand what you've done completely, except for one part.

    How do you know that the function would be continuous at x = 0? The reason I don't get it is because kcos(x) , x < or equal to 0 (that top part). I thought that since one of the equations was equal to 0, that 0 was covered in the graph?

    That's the only part I don't understand.
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  5. #5
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    I suspect that redsoxfan325 knows that this function is continuous at x= 0 because he knows the definition of "continuous"!

    That definition of "f(x) is continuous at x= a" has three parts:
    1) f(a) is defined. That's where you need "0 was covered in the graph".

    2) \lim_{x\to a} f(x) is defined. For a "piecewise defined function" that means, of course, that the two one-sided limits must be the same.

    3) \lim_{x\to a} f(x)= f(a)

    Is that the case for this function? As you say, (1) is true. The two one-sided limits as \lim_{x\to 0^+} f(x)= \lim_{x\to 0} k cos(x) and \lim_{x\to 0^-} f(x)= \lim_{x\to 0} 6e^x- k. (2) What are those limits? Set them equal and solve for k. Of course, you should be able to show that the limit, for that k, is the same as the value of f(x).
    Last edited by mr fantastic; September 17th 2009 at 05:52 AM. Reason: Added a close latex tag
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  6. #6
    Super Member redsoxfan325's Avatar
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    Yeah, sorry about that. I skipped a bunch of preliminary steps.
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  7. #7
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    Okay, thank you for clarifying that.

    So then, applying what you all have said (I hope), let's say you have a problem with three different functions like this:



    You need to see if it's continuous at x = 0 and x = 2, right? So would you do something like,

    e^ (0c) = (0 + c)^2

    and then

    (2 + c)^2 = 2(2) + 6

    and just solve for c?

    I did that and got

    e^(0c)  = (0 + c)^2
    1 = c^2
    c = 1

    and

    (2 + c)^2 = 2(2) + 6
    4 + 4c + c^2 = 6c
    c(4 + c) = 6c
    4 + c = 6
    c = 2

    So basically, c = 1 for the first one, and c = 2 for the second one. Am I doing this correctly? How do I then figure out how to make it continuous at BOTH 0 and 2?


    Sorry I'm asking so many questions on this single thing. I'm just really struggling with this particular concept.
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