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David_Novo
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.
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.
What is the difference?
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
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.
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.
Thanks van1234, any idea about the second question?