For what values of x, if any, does #f(x) = sec((-11pi)/6-7x) # have vertical asymptotes?

Answer 1

#=> x = (-2pin) /7 pm -pi/14 -(11pi)/42 #

# n in ZZ #

We need to consider the definition of:

#sec x = 1/cosx #

Hence:

#sec( (-11pi)/6 -7x ) = 1/( cos( (-11pi)/6 -7x)) #
Hence there are verticle asymtptotes where the denominator #=0#
#=> cos( (-11pi)/6 - 7x) = 0 #
#=> cos ((-11pi)/6 - 7x ) = cos(pi/2) #
Using the general solution for #cosx#:
If #cosx = cos phi # #=> x = 2pin pm phi #
#=>(-11pi)/6 - 7x = 2pin pm pi/2 #
#=> -7x = 2pin pm pi/2 + (11pi)/6 #
#=> x = (-2pin) /7 pm -pi/14 -(11pi)/42 #
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Answer 2

The function f(x) = sec((-11pi)/6-7x) has vertical asymptotes when the value inside the secant function, (-11pi)/6-7x, equals odd multiples of pi/2.

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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.

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.

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.

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.

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