Domain and Range of Inverse Function (1 Viewer)

[cutemouse]

Hello,

I'm stuck on this one... Thanks to anyone who could help me.

State carefully the domain and range of y=cos^-1sqrt((1/4)-x²)

Thanks

 

[Affinity]

Hello,

I'm stuck on this one... Thanks to anyone who could help me.

State carefully the domain and range of y=cos^-1sqrt((1/4)-x²)

Thanks

For that expression to make sense,
sqrt((1/4)-x²) must be between -1 and 1 (so you can take inverse cosines)

this means that (1/4)-x² should be between 0 and 1,
which means that x must lie between -1/2 and 1/2.. and and all values in between are fine, so the domain is [-1/2,1/2]

To find the range you go backwards, x^2 will lie between 0 and 1/4,

so 1/4 - x^2 will also be between 0 and 1/4

so the square toor of that is between 0 and 1/2

so the range of the whole thing is between pi/3 [which is arccos(1/2)]and pi/2 [which is arccos(0)]