# For what values of x, if any, does #f(x) = tan((-7pi)/4-2x) # have vertical asymptotes?

Here, they are given by

See graph: graph{(y+tan(2(x+7/8pi)))=0[-pi pi -pi/2 pi/2]}

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The function f(x) = tan((-7pi)/4-2x) has vertical asymptotes when the tangent function is undefined. The tangent function is undefined when the angle is equal to (2n+1)pi/2, where n is an integer.

To find the values of x that result in vertical asymptotes, we set (-7pi)/4-2x equal to (2n+1)pi/2 and solve for x.

(-7pi)/4-2x = (2n+1)pi/2

Simplifying the equation, we get:

-7pi/4 = (2n+1)pi/2 + 2x

-7pi/4 - (2n+1)pi/2 = 2x

Combining like terms, we have:

(-7pi - 4(2n+1)pi)/4 = 2x

Simplifying further:

(-7pi - 8npi - 4pi)/4 = 2x

(-11pi - 8npi)/4 = 2x

(-11pi - 8npi) = 8x

Solving for x:

x = (-11pi - 8npi)/8

Therefore, the values of x that result in vertical asymptotes for the function f(x) = tan((-7pi)/4-2x) are given by x = (-11pi - 8npi)/8, where n is an integer.

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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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