Inverse (1 Viewer)

[HoldingOn]

http://prntscr.com/kkcoio

Could I get some help with part i, ii, iii, and iv. Thanks

 

[1729]

http://prntscr.com/kkcoio

Could I get some help with part i, ii, iii, and iv. Thanks

http://prntscr.com/kkd19x

idk if (iv) is right, and there also might be a faster way to do (ii)

edit: it's meant to say f'(x) > 0 in (iii)

 

[HoldingOn]

http://prntscr.com/kkd19x

idk if (iv) is right, and there also might be a faster way to do (ii)

edit: it's meant to say f'(x) > 0 in (iii)

Thanks. How do you know by inspection the domain without solving an inequality?

 

[1729]

Thanks. How do you know by inspection the domain without solving inequality?

note the domain of inverse sine is -1 <= x <= 1

if you graph the numerator and the denominator separately you will find that one is always greater than the other and hence the quotient is less than one for all real x.

simplifying the LHS inequality gives you x^2 >= 0 which is also true for all real x.