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johnny101

  • one year ago

Calc, help! How do I determine at what points y=x+2/cos x is continuous?

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  1. Decart
    • one year ago
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    where the function is defined

  2. pgpilot326
    • one year ago
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    what's the def of continuous?

  3. Decart
    • one year ago
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    not at asymptote

  4. johnny101
    • one year ago
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    its not defined in text so yeah im assuming

  5. pgpilot326
    • one year ago
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    Definition in terms of limits of functions The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c).[2] In mathematical notation, this is written as \[ \lim_{x \to c}{f(x)} = f(c).\] In detail this means three conditions: first, f has to be defined at c. Second, the limit on the left hand side of that equation has to exist. Third, the value of this limit must equal f(c).

  6. Decart
    • one year ago
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    as x approaches what value is the function undefined

  7. Decart
    • one year ago
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    take the derivative

  8. johnny101
    • one year ago
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    can you walk me through that step by step if possible? I missed the lecture and am somewhat lost

  9. Decart
    • one year ago
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    you need to use the quotient rule and the derivative of cos is -sin

  10. johnny101
    • one year ago
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    f(x)/g(x)=l/m?

  11. pgpilot326
    • one year ago
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    the function is not defined when \[x=\frac{ \pi }{ 2 }+k\pi\text{, where }k \in \mathbb{Z}\]

  12. pgpilot326
    • one year ago
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    because cos x will be 0 at those values of x and the function will not be deifned there. thus, the function will be discontinuous at those points

  13. johnny101
    • one year ago
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    so what you have above pi/2+ k(pi), how did you determine points off that?

  14. pgpilot326
    • one year ago
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    y = x is continuous for all real x. y = 2/x is continuous for all real x where x is not 0. -1 <= cos x <= 1 for all real x. thus, so long as cos x not = 0, your function will be continuous.

  15. johnny101
    • one year ago
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    ohhhhh. Ok I got it now. Thank you!

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