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13 August, 14:26

Where are the asymptotes of f (x) = tan (4x - π) from x = 0 to x = pi over 2?

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  1. 13 August, 17:02
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    Asymptotes of tan (x) are at (k+1/2) π. (k=any integer)

    So the asymptotes of f (x) are

    f (x) = tan (4x-π ) = tan ((k+1/2) π )

    =>

    4x-π = (k+1/2) π

    = (k+1/2) π +π

    = (k+1/2) π

    =>

    x = (k+1/2) π /4

    = (k/4+1/8) π

    For

    k=-1, x=-π /8 [outside of (0,π /2) ]

    k=0, x=π /8

    k=1, x=3π /8

    k=2, x=5π /8 [outside of (0,π /2) ]

    So the answer is x={π /8, 3π /8}
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