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Danios
High SchoolMathematics 5+3 pts
Where are the asymptotes of f(x) = tan(2x − π) from x = pi over 2 to x = 3 pi over 2?
A) x = 3 pi over 4, x = 5 pi over 4
B) x = 0, x = π, x = 2π
C) x = 0, x = pi over 4
D) x = pi over 2, x = 3 pi over 2
Comments (1) FollowReport by Danios 15.06.2018
hmmm those choices seem to include values outside the provided range, no? I mean, the provided range is [ π/2 , 3π/2 ], only the 2nd and 3rd Quadrants.
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Answers
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Professor1994 Virtuoso
Answer: Option A) x=3π/4, x=5π/4
The asymptotes of the function tan (z) are the values of z that are the odd multiples of π/2:
z=(2n+1)π/2, witn n= ..., -3, -2, -1, 0, 1, 2, 3, ...
In this case z=2x − π, then:
2x − π=(2n+1)π/2
Solving for x: Adding π both sides of the equation:
2x − π+ π=(2n+1)π/2+ π
Adding the terms on the right side of the equation:
2x=[(2n+1)π+2π]/2
Common factor π:
2x=[(2n+1)+2]π/2
2x=(2n+1+2)π/2
2x=(2n+3)π/2
Multiplying both sides of the equation by 1/2:
(1/2)(2x)=(1/2)[(2n+3)π/2]
(2/2)x=(2n+3)π/[(2)(2)]
x=(2n+3)π/4
For n=-1:
x=[2(-1)+3]π/4
x=(-2+3)π/4
x=π/4
x=π/4<π/2 No
For n=0:
x=[2(0)+3]π/4
x=(0+3)π/4
x=3π/4
π/2<x=3π/4<3π/2 Ok
For n=1:
x=[2(1)+3]π/4
x=(2+3)π/4
x=5π/4
π/2<x=5π/4<3π/2 Ok
For n=2:
x=[2(2)+3]π/4
x=(4+3)π/4
x=7π/4
x=7π/4>3π/2 No
Answer: x=3π/4, x=5π/4