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anonymous

  • one year ago

Please help with math!!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    Need an explanation on how to do this.

  3. anonymous
    • one year ago
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    @phi

  4. anonymous
    • one year ago
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    @mathmath333

  5. anonymous
    • one year ago
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    @freckles

  6. anonymous
    • one year ago
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    @imqwerty

  7. anonymous
    • one year ago
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    @triciaal

  8. Owlcoffee
    • one year ago
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    We define continuity for a function "f" as the coicident lateral limit of the intermediate point. We can traduce like this: \[\lim_{x \rightarrow a ^{\pm}} f(x)=f(a)\] What does this mean? In very simple words, the sufficient and necessary condition for a function to be continous at a point as long as the lateral limits of such point have a value coincident with the image of that point. Let's take a look at that and apply it to the problem in question, as you can see. Since the function is divided we will study what happens on that x=-4 point. So we will study the two following limits: \[\lim_{x \rightarrow -4^{-}}\frac{ x+4 }{ x^2-16 }\] and \[\lim_{x \rightarrow -4^{+}}\frac{ c }{ x+12 }\] These two limits arise from the very definition of limits, since when \(x<-4\), the function f takes the structure of \(\frac{ x+4 }{ x^2-16 }\) and when \(x \ge -4\) the function f takes the structure \(\frac{ c }{ x+12 }\). Now, solving those limits will give a value, and then you will want to use "c" as a variable to make them equal, therefore making the function continous.

  9. anonymous
    • one year ago
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    But when you plug in -4 they equal 0 so how can you equal both equations together?

  10. anonymous
    • one year ago
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    @Owlcoffee

  11. Owlcoffee
    • one year ago
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    Let's first work with: \[\lim_{x \rightarrow (-4)^-} \frac{ x+4 }{ x^2-16 }=\frac{ 0 }{ 0 }\] This is an indetermination, which means that that you'll have to apply some factorization strategies in order to lift up the indetermination and solve the limit.

  12. anonymous
    • one year ago
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    So then it would be \[\frac{ 1 }{ (x-4) }\] right?

  13. anonymous
    • one year ago
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    @Owlcoffee

  14. Owlcoffee
    • one year ago
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    Yes, correct, and then you plug "-4", what do you get?

  15. anonymous
    • one year ago
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    -1/8

  16. anonymous
    • one year ago
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    Then what do I do?

  17. Owlcoffee
    • one year ago
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    Good, now solve the second limit: \[\lim_{x \rightarrow (-4)^+}\frac{ c }{ x+12 }\] In this case, just treat "c" as a constant and apply the limit.

  18. anonymous
    • one year ago
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    \[\frac{ -1 }{ 8 }=\frac{ c }{ 8 }\]\[-8=8c\]\[-1=c\]

  19. anonymous
    • one year ago
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    Is that right?

  20. Owlcoffee
    • one year ago
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    It is correct.

  21. anonymous
    • one year ago
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    Okay thanks!

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