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Help needed with Domian and Range of inverse sine functions (2 Viewers)

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 :|

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