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1 August, 22:35

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

x = pi over 4, x = 3 pi over 4

x = 0, x = pi over 4

x = pi over 2, x = 3 pi over 2

x = 3 pi over 8, x = 5 pi over 8

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  1. 2 August, 00:47
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    Thus the asymptotes are π/8 and 3π/8

    Step-by-step explanation:

    An asymptote is a line that a graph approaches without touching. The asymptotes are where the graph is undefined.

    tan (x) = sin (x) / cos (x), where cos (4x-π) = 0

    cos (4x-π) = 0 when inside is - π/2, π/2, 3π/2

    4x-π = π/2

    Add π at both sides:

    4x-π+π = π/2 + π

    4x = π+2π/2

    4x = 3π/2

    x = 3π/8

    4x-π = 3π/2

    4x = 3π/2 + π

    4x = 5π/2

    x = 5π/8

    This one is outside the interval.

    4x-π = - π/2

    4x = - π/2 + π

    4x = - π+2π/2

    4x=π/2

    x = π/8

    Thus the asymptotes are π/8 and 3π/8 ...
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