New posts

Best of luck to the class of 2025 for their HSC exams. You got this! Let us know your thoughts on the HSC exams here

Help needed with Domian and Range of inverse sine functions (1 Viewer)

Thread starter blackops23
Start date Jan 16, 2011

B

blackops23

Member

#21

Trebla said:
Since

$0<\dfrac{1}{x^2+1}\leq 1\,$ for all real $x$

then

$\sin^{-1}0<\sin^{-1}\left(\dfrac{1}{x^2+1}\right)\leq \sin^{-1}1\\\\\Rightarrow 0<\sin^{-1}\left(\dfrac{1}{x^2+1}\right)\leq \dfrac{\pi}{2}$

Thankyou

H

hscishard

Active Member

#22

Note the < sign and not <= sign next to the zer0 part.
Sicne y=0 is the horizontal asymptote of 1/x^2+1

Last edited: Jan 17, 2011

B

Bored Of Fail 2

Banned

#23

hscishard said:
Note the < sign and not <= sign next to the zer0 part.
Sicne y=0 is the horizontal asymptote of 1/x^2+1

lol yeh I missed that :|

You must log in or register to reply here.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top