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anonymous

  • 5 years ago

Prove that the graph is continuous or not continuous at x = 2.

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  1. anonymous
    • 5 years ago
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    i have attached the graph

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  2. amistre64
    • 5 years ago
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    its an empty graph

  3. anonymous
    • 5 years ago
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    one sec

  4. amistre64
    • 5 years ago
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    and a picture is better, since not everyone has microsoft office

  5. anonymous
    • 5 years ago
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    sure one sec

  6. anonymous
    • 5 years ago
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    there the pic is attached

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  7. anonymous
    • 5 years ago
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    Or even if they do have office, opening docs from unknown sources can sometimes be problematic.

  8. anonymous
    • 5 years ago
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    f(2) exists (infact f(2) = 4

  9. amistre64
    • 5 years ago
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    to prove: the lefthand limit has to equal the right hand limit at x = 2

  10. amistre64
    • 5 years ago
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    how to prive that without actual functions? my best guess is just to point to it on the graph and say, "see! right there"

  11. anonymous
    • 5 years ago
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    do you have pdf

  12. amistre64
    • 5 years ago
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    no

  13. anonymous
    • 5 years ago
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    i can send you a similar graph which the teacher gave us to review with

  14. amistre64
    • 5 years ago
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    i got the picture of the graph now; its just that there is no peicewise function defining the curves

  15. anonymous
    • 5 years ago
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    You can see what the left and right hand limits are by looking at the graph.

  16. anonymous
    • 5 years ago
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    And they aren't both f(-2)

  17. anonymous
    • 5 years ago
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    err f(2)

  18. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow 2^+}f (x)=5\neq \lim_{x \rightarrow 2^-}f(x)=2\] Hence f is not continuous at \(x=2\).

  19. anonymous
    • 5 years ago
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    k thats one

  20. anonymous
    • 5 years ago
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    A function is continuous about a point p if and only if p in in the domain of f, and the limit from the left = the limit from the right = f(p)

  21. anonymous
    • 5 years ago
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    f(2) exists (in fact f(2) = 4 is this correct

  22. amistre64
    • 5 years ago
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    sprinkle in some epsilons and deltas for good effect :)

  23. anonymous
    • 5 years ago
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    one second i will attach the sample graph she provided i think thats how she wants the answers

  24. amistre64
    • 5 years ago
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    we can see that there for every epsilon in the neighborhood of L can be produced by a delta such that 0<|x-c|<d in the neightborhood of 2

  25. anonymous
    • 5 years ago
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    forget this problem adding a new one

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