The global extreme values of a function in a set #[a,b]# are to be searched between the local extreme values in the set #[a,b]# and in #a# or #b#.
The domain of the function is #D=[-2,2]#, because the radical argument has to be not-negative.
Let's search the local extreme:
#y'=1*sqrt(4-t^2)+t*1/(2sqrt(4-t^2))*(-2t)=#
#=sqrt(4-t^2)-t^2/sqrt(4-t^2)=(4-t^2-t^2)/sqrt(4-t^2)=2*(2-t^2)/sqrt(4-t^2)#.
Now:
#y'>=0#
if
#2-t^2>=0# (the radical is positive or zero in its domain!),
#-sqrt2<=t<=sqrt2#.
#f(-sqrt2)=-sqrt2*sqrt(4-2)=-2#
#f(sqrt2)=sqrt2*sqrt(4-2)=2#.
So the point #A(-sqrt2,-2)# is a local minimum and it is not in the set #[-1,2]# and the point #B(sqrt2,2)# is a local maximum that is in the set.
#f(-1)=-sqrt3#
#f(2)=0#.
SO:
The point #B(sqrt2,2)# is the global maximum and the point #C(-1,-sqrt3)# is the global minumum.
graph{xsqrt(4-x^2) [-10, 10, -5,5]}
