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anonymous

  • one year ago

What are the domain of these two functions:

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  1. anonymous
    • one year ago
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    \[y=-3\sec(\pi-2x)+5\]

  2. anonymous
    • one year ago
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    \[y=-2\sin(5/4x)\]

  3. anonymous
    • one year ago
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    @Nnesha

  4. anonymous
    • one year ago
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    somebody please help

  5. anonymous
    • one year ago
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    @Vocaloid

  6. anonymous
    • one year ago
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    Domain represents the value for x thus in the second equation is there any value that x cannot be?

  7. anonymous
    • one year ago
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    the sine curve has its domain unrestricted where as the secant function has vertical asymptotes, hence a restricted domain.

  8. anonymous
    • one year ago
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    therefore the domain for the first function is (-infinity,infinity)

  9. anonymous
    • one year ago
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    but I do not know the domain for the second function, what is it? @Deeezzzz

  10. anonymous
    • one year ago
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    Are you familiar with finding the period of the secant func?

  11. anonymous
    • one year ago
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    yes

  12. anonymous
    • one year ago
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    The period of the second function is 2

  13. anonymous
    • one year ago
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    @Deeezzzz

  14. mathstudent55
    • one year ago
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    The first function uses the secant function. Remember this identity. \(\sec \theta = \dfrac{1}{\cos \theta}\) The secant is not defined where the cosine equals zero.

  15. mathstudent55
    • one year ago
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    In the second function, you have the sine function. The sine function is defined for every value of theta.

  16. anonymous
    • one year ago
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    wait, so for the second function the domain is (-infinity,infinity)

  17. mathstudent55
    • one year ago
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    yes

  18. anonymous
    • one year ago
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    I still dont get the domain for the first function

  19. mathstudent55
    • one year ago
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    There is no restriction on the domain of the sine function.

  20. anonymous
    • one year ago
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    how would i write it down in terms of interval notation

  21. mathstudent55
    • one year ago
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    Where is the cosine equal to zero?

  22. anonymous
    • one year ago
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    its undefined?

  23. anonymous
    • one year ago
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    @mathstudent55

  24. mathstudent55
    • one year ago
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    The figure below shows where the cosine equals zero. It is at those points where the secant is undefined. |dw:1438040478384:dw|

  25. mathstudent55
    • one year ago
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    The first function has a secant, so its domain is all reals except for integer multiples of \(\dfrac{\pi}{2} \).

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