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For what values of x, if any, does #f(x) = cot((7pi)/12-x) # have vertical asymptotes?

Answer 1

Isaiah Allen

#x= (7pi)/12 - 2pin #
#x = -(5pi)/12 - 2pin#

For #n in ZZ#

To approach this we must cosnider how #cotx = cosx/sinx#

Hence for #cotx# to have verticle asymptotes, the denominator must be 0, so hence #sinx = 0#

So hence in this circumstance ; #sin((7pi)/12 - x) = 0# for the verticle asymptotes

So hence now we consider the general solution of; #sinx = sina# Then #x = 2pin + a# and #x = 2pin + pi -a#

So hence #sin( (7pi) /12 - x) = sin0#

Hence; #(7pi)/12 - x = 2pin# and #(7pi)/12 - x =2pin + pi#

Hence rearanging to yeild;

#x= (7pi)/12 - 2pin # #x = -(5pi)/12 - 2pin#

For #n in ZZ#

We can consider this by graphing; graph{cot( 7pi/12 -x) [-4.354, 5.644, -2.28, 2.72]}

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Answer 2

Emma Aldridge

The function f(x) = cot((7pi)/12-x) has vertical asymptotes at x = (7pi)/12 + n*pi, where n is an integer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.