How do you find the domain and range of f(x) = ( tan(2x) ) / ( (sin^-1)(x)- pi/3)f(x)=tan(2x)(sin1)(x)π3?

1 Answer
Aug 6, 2018

Domain: x in [ - pi/2, 1/2sqrt 3 ) U ( 1/2sqrt3, - pi/4)x[π2,123)U(123,π4)
U (- pi/4, pi/4 ) U ( pi/4, pi/2]U(π4,π4)U(π4,π2]
Range: ( - oo, oo )#

Explanation:

The presence of sin^(-1)xsin1x restricts the domain to be

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

y = (tan 2x)/(sin^(-1)x - pi/3 ), x ney=tan2xsin1xπ3,x asymptotic 1/2sqrt3, +-pi/4#.

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

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