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anonymous

  • one year ago

use the definition of continuity to determine whether f is continuous at a.

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  1. anonymous
    • one year ago
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  2. perl
    • one year ago
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    I can't read that because i don't have microsoft word. one moment

  3. anonymous
    • one year ago
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    ok

  4. perl
    • one year ago
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    Dowloading openoffice

  5. anonymous
    • one year ago
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    ok thank you so much for helping me

  6. perl
    • one year ago
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    apparently the file to download is huge

  7. anonymous
    • one year ago
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    f(x)= { 2-x if x<1 { 1 if x=1 { x^2 if x>1 a=1

  8. perl
    • one year ago
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    for continuity we need to check the points where the function changes, the rest of the function is continuous in its respective parts.

  9. anonymous
    • one year ago
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    ok how do we do that

  10. perl
    • one year ago
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    Definition of continuity: Left hand limit = right hand limit = value at the point

  11. perl
    • one year ago
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    We can write this in symbols.

  12. anonymous
    • one year ago
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    you write it with a little apostrophe thing right

  13. perl
    • one year ago
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    Definition of continuity: $$ \Large \lim_{x \to a} f(x) = f(a)$$But this implies $$ \Large \lim_{x \to a^{+}} f(x) = \Large \lim_{x \to a^{-}} f(x) = f(a) $$

  14. perl
    • one year ago
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    So we need to check the left limit and right limits and make sure it equals to the value at that point.

  15. perl
    • one year ago
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    $$ \rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} } $$

  16. anonymous
    • one year ago
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    for all the x's i replace them with a 1?

  17. perl
    • one year ago
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    \[ \rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} } \] \[ \Large \lim_{x \to 1^{+}} f(x) = \\~\\ \Large \lim_{x \to 1^{-}} f(x) = \\~\\ \Large f(1) = \]

  18. perl
    • one year ago
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    yes thats correct. and we need to check that all three of those are equal . thats the continuity condition

  19. anonymous
    • one year ago
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    so the answer is 1 for all of them.

  20. perl
    • one year ago
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    \[ \rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} } \] \[ \Large \lim_{x \to 1^{-}} f(x) =\lim_{x \to 1^{-}} 2-x = 2-1 = 1 \\~\\ \Large \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} x^2 = 1^2 = 1 \\~\\ \Large f(1) = 1 \]

  21. anonymous
    • one year ago
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    Thank You So Much,can you help me with one more similar to this one.

  22. perl
    • one year ago
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    ok

  23. anonymous
    • one year ago
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    determine for what numbers, if any, the given function is discontinuous. f(x)={x+7 if x is less than or equal to 0 {7 if 0 is <x less than or equal to 3 {x^2-1 if x>3

  24. anonymous
    • one year ago
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    ok whats the difference between the first set and the second set. (not the questions but what you did)

  25. perl
    • one year ago
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    there are two potential points of discontinuity, where the function abruptly changes. it changes at x =0 and x = 3. so we have to test those two parts independently

  26. anonymous
    • one year ago
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    ok so how do we do that

  27. perl
    • one year ago
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    We can look at the graph that might make it more clear https://www.desmos.com/calculator/pcjrrf5xdh

  28. perl
    • one year ago
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    edit* \[ \rm \Large{f(x)= \begin{cases} x+7 ~~~if~ x\leq0\\ 7 ~~~~~~~~ ~\rm ~if~ 0 < x\leq 3\\ x^2-1 \rm ~~if~ x>3 \end{cases} } \] \[ \Large \lim_{x \to 0^{-}} f(x) =\lim_{x \to 0^{-}} x+7 = 0+7 = 7 \\~\\ \Large \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} 7 = 7 \\~\\ \Large f(0) =0+7 = 7 \] \[ \Large \lim_{x \to 3^{-}} f(x) =\lim_{x \to 3^{-}}7 = 7 \\~\\ \Large \lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} x^2-1 = 3^2-1 = 8 \\~\\ \Large f(3) = 7 \]

  29. perl
    • one year ago
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    from the graph and from the continuity definition we see that there is a point of discontinuity at x = 3. x=0 is fine

  30. anonymous
    • one year ago
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    so all the question was asking for was two points

  31. perl
    • one year ago
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    Before we do any work the function is potentially discontinuous at the two points x=0 and x = 3 because the function changes suddenly at those points. But after analyzing the function graphically, or using the limit above, we see that it is actually only discontinuous at one point. x= 3

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