How do you find the domain and range of $f(x) = ( tan(2x) ) / ( (sin^-1)(x)- pi/3)$ ?

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Answer 1 · 2018-08-06T07:27:14

Domain: $x in [ - pi/2, 1/2sqrt 3 ) U ( 1/2sqrt3, - pi/4)$
$U (- pi/4, pi/4 ) U ( pi/4, pi/2]$
Range: ( - oo, oo )#

The presence of $sin^(-1)x$ restricts the domain to be

( - pi/2, pi/2 ). sans asymptotic x, within.

$y = (tan 2x)/(sin^(-1)x - pi/3 ), x ne$ asymptotic 1/2sqrt3, +-pi/4#.

Range: $y in ( - oo, oo )$

See graph.
graph{(y( arcsin (x ) - pi/3 ) - tan (2x ))(x-1/2sqrt3-0.0001y) = 0[-6 6 -3 3]}

How do you find the domain and range of f(x) = ( tan(2x) ) / ( (sin^-1)(x)- pi/3)f(x) = ( tan(2x) ) / ( (sin^-1)(x)- pi/3)?