iheartcookies1012
Member
- Joined
- Jul 7, 2023
- Messages
- 39
- Gender
- Female
- HSC
- 2024
how do u find the domain and range of y=sin(2cos^-1 x)?
wait lmao my bad i must have swapped the two ill edit itwait but for sin^-1(sqrtx) there is a restriction on range... its not [0,pi/2] instead of [-pi/2, pi/2], is this becus the domain got restricted so the range also changed?
yep it does thanks !! gl w english (i hate it too)can u read and see if that makes sense or works
im doing english rn and my draft is due at 11.59pm so i dont have much time to check : p
What I'd recommend u do is let u=f(x) in y=sin^-1[f(x)] and using this information, sketch both graphs, i.e: y=sin^-1(u) and u=f(x).wait but for sin^-1(sqrtx) there is a restriction on range... its [0,pi/2] instead of [-pi/2, pi/2], is this becus the domain got restricted so the range also changed?
true composite functions is the best way to set these up
wait hold up where'd I go wrong>?true composite functions is the best way to set these up
btw ur graph for the second one is incorrect tho
the domainwait hold up where'd I go wrong>?
Oh nvm I understand now lmfao. (mbmb)the domain
x^2 is positive but that doesn't mean x has to be
negative x values just become positive so the graph exists for [-1,0] as well
it would be a horizontally mirrored version of the part on the right side of the graph