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Math Help - Continuity

  1. #1
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    Continuity

    Hey,

    I'm a little confused on the continuity rules.

    This is the equation of the graph: (supposed to be one of those split graph things but the format got screwed up)

    f(x) = ( x+2 < 0), ( e^x + 1 <=x<2), (x^2 + 1 x>2)


    Continuity rules:

    I don't understand the second and third rules of continuity, can you explain them to me PLEase (and show me how to do it for my question (greatly appreciated))!!!!!!!!


    Thanks
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Hello!!

    This appears to be three different functions:

    f(x) = (x+2), for x < 0

    f(x) = e^{x+1}, for x < 2

    f(x) = x^2, for x > 2

    Is that what you have? Not sure what your equation is. Can you clarify that?

    For the continuity rules #2 & #3:

    #2 says that as the x values on the graph of f(x) get closer and closer to c, there is a guaranteed y value associated with each of those x values

    #3 says that if you evaluate f(x) at c, meaning you plug the the value of c into the x values of your function, then the answer you get will be the limit of your function.

    Hope that helps!

    #3 says
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  3. #3
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    OO yeah sorry, I didn't put the limit I'm finding in my last post, I'mdoing as x approaches 0

    Hey, Uhhh... the way the graph is supposed to be is that values of x less than 0 will look like a line (because of the equation x+2) as you move on the graph changes because the equation x+2 no longer applies the values of x greater than 0 so you use the other equations.


    So for rule #2 does my equation work because there is no value associated with two.

    Also whether I'm missing 0 does not matter right (0 is the limit I'm calculating)

    #3 So for rule number three f(0) is supposed to equal to 0 right???

    thanks
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  4. #4
    Senior Member mollymcf2009's Avatar
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    f(x) = ( x+2 < 0), ( e^x + 1 <=x<2), (x^2 + 1 x>2)

    Ok, so for the above, does " x+2 < 0" mean

    \lim_{x\to0} x+2

    and

    " ( e^x + 1 <=x<2)" means
    \lim_{x\to2} e^x +1

    and

    "(x^2 + 1 x>2)" means
    \lim_{x\to2} x^2 +1

    Am I reading that correctly?
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  5. #5
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    Uhhh.... you haven't quite got it. I wrote this long explanation to explain it, it's supposed to be a "piecewise defined function" but I just figured, hey Let's just change the question just so I can get the concept right.

    let's say my question asks for the limit of x+2 as x approaches 0

    so while following the three rules I say
    1) does y exist at x=0? and f(0) = 0+2 so I say Yes
    2) the limit of x+2 as x approaches zero the same from both the right and the left? --> I say yes and put test values in to make sure both sides approach 2
    3) This one I'm a little iffy on ----> so: do I say that f(0) is supposed to equal 0??? but it doesn't, and I already know ahead of time that a line (x+2) is a continuous function!!!!! --> a little lost sorry!!!!





    LOng Explanation Not Necessary to read

    oops I just noticed a zero is missing in one of them that might be what is confusing about it:

    f(x) = ( x+2 < 0), ( e^x + 1 0<=x<2), (x^2 + 1 x>2)

    not sure if this makes more sense now but my text book calls it piecewise defined functions.

    Umm... all of the equations up there correspond to different x values. So the Y value I get from an x that is less than 0 will correspond to only (x+2). The Y value I get from a value of lets say 1 will only correspond to (( e^x + 1 0<=x<2)

    so you can imagine that the graph will look kind of warped in that when x is less than 0 the graph looks like a line but as you move to 0 the graph starts turning into a curve (because of e^x + 1, exponential graph) and then a curve again at (x^2 + 1, parabola) at x>2.

    and then the question asks me to find the limit of 0

    So if I'm finding 0 from the left I would use the x+2 equation (because it's less than 0) and from the right I would use e^x + 1 (because it's greater than 0 and moves towards zero)
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  6. #6
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    Quote Originally Posted by Mudd_101 View Post
    Hey,

    I'm a little confused on the continuity rules.

    This is the equation of the graph: (supposed to be one of those split graph things but the format got screwed up)

    f(x) = ( x+2 < 0), ( e^x + 1 <=x<2), (x^2 + 1 x>2)


    Continuity rules:

    I don't understand the second and third rules of continuity, can you explain them to me PLEase (and show me how to do it for my question (greatly appreciated))!!!!!!!!


    Thanks
    The function is f(x) = \left\{ \begin{array}{ll}<br />
x + 2, & x < 0 \\<br />
e^x + 1, & 0 \leq x < 2 \\<br />
x^2, & x > 2 \end{array} \right.

    The only potential continuity trouble is at x = 0 and x = 2.

    Note that:

    1. \lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} (x + 2) = 2.

    2. \lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} (e^x + 1) = 2.

    Therefore \lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = 2 and so \lim_{x \rightarrow 0} f(x) = 2.

    3. f(0) = e^0 + 1 = 2.

    Therefore \lim_{x \rightarrow 0} f(x) = f(0) and so the function is continuous at x = 0.

    Now you need to apply the same thinking to x = 2. You will find that \lim_{x \rightarrow 2^-} f(x) \neq \lim_{x \rightarrow 2^+} f(x) and so \lim_{x \rightarrow 2} f(x) does not exist. So the function is not continuous at x = 2.

    All of this is very obvious of course if you draw a graph of the function .....
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  7. #7
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    Hey, Okay, yeah I agree it's easy if I draw a graph but my teacher says... that I have to use the three rules so:

    that being said, the three rules would be

    1) the left and the right equal one another
    2) f(c) exists
    3) f(c) must equal to what I got in the first part?????????

    I'm still a little confused on the last one Please clarify, Thanks , greatly appreciated

    @!@@
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  8. #8
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    Quote Originally Posted by Mudd_101 View Post
    Hey Mr. Fantastic the post was useful and I agree it is obvious if I draw a graph however@! On my test my teacher says I have to prove it using the three rules of continuity.

    You showed me how to do it using the rule that says the left side must equal the right can yOu please show me how to do the other rules:

    the two I'm comfortable with are

    1) the approach from the left should equal the approach from the right
    2) F(c) (c being the limit number) must exist

    I don't get the last one Help MePlease!!!!!!!
    I have used the last one:

    Quote Originally Posted by Mr Fantastic
    [snip]
    Therefore and so the function is continuous at .
    [snip]
    Read my reply more carefully and you will see that I have used all three of your the 'rules'.
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  9. #9
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    Uhhhh.... Okay, I think I get it, Thanks.

    :S:S:S:S:S:S: :S
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