tacogym27101990
Member
can someone tell me the horizontal asymptotes of y=1/arctanx?
thanks
thanks
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- Thread starter tacogym27101990
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Slidey
But pieces of what?
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You can find them yourself this way:
y=tan(x)=sin(x)/cos(x). Asymptotes at cos(x)=0, which is x=pi/2+2kpi. For inverse functions, rotate the asymptotes 90 degrees, thus the asymptotes for y=arctan(x) are y=pi/2+2kpi. Restrict range to ensure it's a function (rather than relation): y=pi/2 and y=-pi/2.
Thus the horizontal asymptotes are y=1/pi/2=2/pi and y=-1/pi/2=-2/pi
What about vertical asymptotes? None existed before we flipped the function, but we know that 1/0 causes problems. Let's see what values of the denominator give 0: arctan(x)=0, x=0. Test either side of x to determine that x=0 is a vertical asymptote.
y=tan(x)=sin(x)/cos(x). Asymptotes at cos(x)=0, which is x=pi/2+2kpi. For inverse functions, rotate the asymptotes 90 degrees, thus the asymptotes for y=arctan(x) are y=pi/2+2kpi. Restrict range to ensure it's a function (rather than relation): y=pi/2 and y=-pi/2.
Thus the horizontal asymptotes are y=1/pi/2=2/pi and y=-1/pi/2=-2/pi
What about vertical asymptotes? None existed before we flipped the function, but we know that 1/0 causes problems. Let's see what values of the denominator give 0: arctan(x)=0, x=0. Test either side of x to determine that x=0 is a vertical asymptote.