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How do you find the domain and range of #y=ln(tan^2(x))#?

Answer 1

Asymptotic #x ne k pi darr and ne (2k + 1)pi/2 uarr, k = 0, +-1, +-2, +-3, ...#

#y = ln ((tan x )^2) = 2 ln tan x# is real when

#x ne k pi , ne (2k + 1)pi/2 and notin Q_2 # or #Q_4#,

to negate respectively, #sin x = 0 and cos x = 0 #. Graph: graph{(tan x - e^(y/2))(x)(x^2-1/4(pi)^2)(x^2-(pi)^2)(x^2-9/4(pi)^2)(x^2-4(pi)^2)(x^2-25/4(pi)^2)(x^2-9(pi)^2)=0}

Note that #y to oo, x to ( 2k + 1 )pi/2 and - oo, x to kpi#.

It is delightful yo observe alternate asymptoticity, by sliding the graph #uarr and darr#. As you just see, it is in hiding.

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Answer 2

Adam Aguirre

To find the domain and range of ( y = \ln(\tan^2(x)) ):

Domain:
- ( \tan(x) ) is defined for all real numbers except odd multiples of ( \frac{\pi}{2} ) (i.e., ( x ) cannot be ( \frac{(2n+1)\pi}{2} ) where ( n ) is an integer).
- ( \tan^2(x) ) is always non-negative since it squares the tangent function.
- ( \ln(u) ) is defined only for positive ( u ).
- Therefore, the domain of ( y = \ln(\tan^2(x)) ) is ( x ) such that ( x ) is not an odd multiple of ( \frac{\pi}{2} ).
Range:
- The range of ( y = \ln(\tan^2(x)) ) is all real numbers since ( \tan^2(x) ) can take any non-negative value, and ( \ln(u) ) is defined for all positive ( u ).

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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.