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mathcalculus Group Title

Help: CONTINUITY and removable discontinuity: (attached below)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. RogerSmith Group Title
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    wouldn't the limit from the left be equal to 40, not 4

    • one year ago
  3. mathcalculus Group Title
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    yes, it is 40. i sketched the graph. now i didn't see 40 in it.

    • one year ago
  4. mathcalculus Group Title
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    every graph that i sketched i usually can tell the answer. BUT for this one, i wasn't able to.

    • one year ago
  5. mathcalculus Group Title
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    can you tell me whats wrong? where did i go wrong in the problem?

    • one year ago
  6. mathcalculus Group Title
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    @RogerSmith

    • one year ago
  7. RogerSmith Group Title
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    okay when you look at the piecewise function, you are first trying to see if the function is continuous. You can see that it is because if you look at the limit you have at the top (steps 1-3), you will see that when you approach a from the left, you must use the equation 8x-8. If you are approaching the point from the right, you will use the other equation (1/x+4)

    • one year ago
  8. RogerSmith Group Title
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    now the point in question is 6. It is asking you whether the function is continuous at a=6. That means if we take the limit from the left side, and the limit from the right side, they should both equal to the same number. Now you have the part correct when you approach 6 from the right. If you look at the piecewise function it says when a is GREATER THAN OR EQUAL TO 6, use the equation on the bottom. You correctly solved this by plugging in 6, and you see that your answer for the right hand limit is 1/10

    • one year ago
  9. RogerSmith Group Title
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    Now, since we know that when we take the limit to the right of 6, you should be approaching the value of 1/10. however, check the left hand limit. if we are approaching 6 from the left, that means we are chosing values smaller than 6. If you look at the piecewise again, it says for values <6, use 8x-8. Plug 6 in for X and solve. 8(6)-8 = 48-8 = 40

    • one year ago
  10. RogerSmith Group Title
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    The left and right hand limits do not equal the same thing. you can see that from the left, x approaches 40, while from the right, it approaches 1/10. Since the two don't match, there is a jump discontinuity. Therefore, the function is not continuous at x=6.

    • one year ago
  11. RogerSmith Group Title
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    f(6) = 1/10 f(6) from the right =approaches 1/10 f(6) from left = approaches 40

    • one year ago
  12. mathcalculus Group Title
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    okay i figured out from the right side. 1/10 but when i sketched the graph i couldn't tell where x appraoches to 6 from the left side

    • one year ago
  13. mathcalculus Group Title
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    normally when i sketch i can tell, why is it that i couldn't see it before?

    • one year ago
  14. mathcalculus Group Title
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    @RogerSmith

    • one year ago
  15. mathcalculus Group Title
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    i see your point but i can't always plug in any number into the first function. it wouuld be wrong. in this case, we can.

    • one year ago
  16. mathcalculus Group Title
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    however, i want to know how to get 40 from the sketch.

    • one year ago
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