IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

theshreadder

Find the domain and range of the fn

f(x)=sin^-1{log₂(1/2 x²)}

him26.89

look,

x is not 0 due to log.

for domain :-

-1

log₂(x²)/2

1

1)2^-1

x²/2

=> 0

x² - 1.

2)x²/2

2

x² - 4

0

x belongs to [-2,-1]+[1,2]

range:-

f(X) is [-pie/2,pie/2].

extreme values are attained at x =1 &2

magiclko

f(t)=sin^-1t

is defined only if t

[-1,1]

that means {log₂(1/2 x²)}

[-1,1]

=> 2^-1

x²/2 and x²/2

2

=> 0

x² - 1. and x² - 4

0

=> x

(-1,1) and x

[-2,2]

combining both, [-2,-1]

[1,2]

manisha_bhandari

how one can know extreme values of range ofa function