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F(x) = sin(-3x) increase decrease intervals

  1. Mar 6, 2015 #1
    Hi everyone.

    Could anyone help me solve this. I need to find the increase and decrease intervals of f(x) = sin(-3x).

    The increase intervals of sin(x) are [itex]-\pi/2 + 2\pi*k < x < \pi/2 + 2\pi*k[/itex]

    Is the following the right way to solve my problem (for increase intervals)? [itex]-\pi/2 + 2\pi*k < -3x < \pi/2 + 2\pi*k[/itex]
    By multiplying by -3 you get: [itex]-\pi/6 + 2/3\pi*k < x < \pi/6 + 2/3\pi*k[/itex] (1)

    Although when plotting sin(-3x) this interval (1) is the decrease one. Why?
  2. jcsd
  3. Mar 6, 2015 #2


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    sin(-3x)=-sin(3x). Work with this form, since you don't have to work with things going in opposite directions.
  4. Mar 6, 2015 #3
    How would you use sin(x) = -sin(x) in this problem?
  5. Mar 7, 2015 #4
    If sin(x) is increasing then -sin(x) is decreasing, vice versa.
  6. Mar 7, 2015 #5
    Yes, thank you. I have already solved it this way. My textbook had an error in the solutions sheet. They just solved the double inequality and did not exchange the increase and decrease intervals.
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