Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2 ?
Question
Answer:
[tex]tan(4x- \pi )= \frac{sin(4x- \pi )}{cos(4x- \pi )} [/tex]The asymptotes are where the graph is undefined. Since: tan(x) =sin(x)/cos(x)
It is where cos(4x-π) = 0
cos(4x-π) = 0 when the inside is -π/2 , π/2 , 3π/2
4x - π = π/2
4x = π/2 + π
4x = 3π/2
x = 3π/8
4x - π = 3π/2
4x = 3π/2 + π
4x = 5π/2
x = 5π/8
This ones outside the interval (5π/8 > π/2) , try -π/2
4x - π = -π/2
4x = -π/2 + π
4x = π/2
x = π/8
Asymptotes are π/8 and 3π/8
solved
general
10 months ago
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