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

Kathy Robinson

in Mathematics

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Kathy Robinson on March 21, 2019

Asymptotes of tan(x) in k (k+1/2)π. (k=any integer), So that the asymptotes of f(x) aref(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)n k=-1, x=-π/8 [out of (0,π/2)]k=0, x=π/8k=1, x=3π/8k=2, x=5π/8 [out of (0,π/2)]So the answer is x={π/8, 3π/8}


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