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David_Novo

  • 3 years ago

1- Can anyone tell me how to solve the limits required for question 3.a Section D Problem Set 1 6 tan(3x)(sec(3x)^2) = 6 sin (x)/ (cos(x)^3) 2 - How do the 3's before the x's cut? Thanks in advance.

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  1. calculusfunctions
    • 3 years ago
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    Is your question: 6tan(3x)sec²(3x) = (6sinx)/cos³x or 6tan(3x)sec(3x)² = (6sinx)/cos(x)³ ? Because there is fundamental difference between those two equations. When you don't express yourself clearly, it makes it difficult for people to help you.

  2. David_Novo
    • 3 years ago
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    What is the difference?

  3. van1234
    • 3 years ago
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    He is asking if you are cubing/squaring the result you get after taking the cos of the angle or are you taking the cos of the cube of the angle. (I am pretty sure hes asking: 6tan(3x)sec²(3x) = (6sinx)/cos³x

  4. David_Novo
    • 3 years ago
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    Yes, that's the equation van1234. The cube is outside the cos function, as I wrote. Any ideas how the 3's inside the tan and sec functions cut? Btw any insight on the first question? Thanks.

  5. van1234
    • 3 years ago
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    Is this the question you are referring to in your first question: \[(x-2)\div(x ^{2}-4)\] If so, then here's my solution. You know that the denominator is a difference of squares, so you can factor that to be: (x+2) (x-2) Then you see that the x-2 's cancel, (keep in mind though that you will have a hole at x = 2, as x=2 still does not work in the original function). You are left with: \[1 \div (x +2)\] Which means that there is an asymptote at x = -2.

  6. David_Novo
    • 3 years ago
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    Thanks van1234, any idea about the second question?

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