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anonymous

  • 2 years ago

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

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  1. anonymous
    • 2 years ago
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    where the function is defined

  2. anonymous
    • 2 years ago
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    what's the def of continuous?

  3. anonymous
    • 2 years ago
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    not at asymptote

  4. anonymous
    • 2 years ago
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    its not defined in text so yeah im assuming

  5. anonymous
    • 2 years 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. anonymous
    • 2 years ago
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    as x approaches what value is the function undefined

  7. anonymous
    • 2 years ago
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    take the derivative

  8. anonymous
    • 2 years 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. anonymous
    • 2 years ago
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    you need to use the quotient rule and the derivative of cos is -sin

  10. anonymous
    • 2 years ago
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    f(x)/g(x)=l/m?

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

  12. anonymous
    • 2 years 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. anonymous
    • 2 years ago
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    so what you have above pi/2+ k(pi), how did you determine points off that?

  14. anonymous
    • 2 years 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. anonymous
    • 2 years ago
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    ohhhhh. Ok I got it now. Thank you!

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