How do you find the global extreme values for #f(t) = tsqrt(4 - t²)# on [-1,2]?

1 Answer
Jul 1, 2015

Answer:

Maximum is #(sqrt2,2)# and minimum is #(-1,-sqrt3)#.

Explanation:

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]}