Absolute Min of y=x Sqr.Root (3-x) in Domain (-infinity, 3]

[mattmannmf]

fuction is y=x sqr.root (3-x)

Domain (-infinity, 3]
i got my critical numbers to be x= 2, 3

Now i know the absolute max is x=2.

but for the absolute min, would it be at x=3? or none since the function continues to go to negative infinity as x approaches negative infinity?

 

[Quantumpencil]

Well, assuming that the function goes to negative infinity, then there is no lower bound. It contradicts the definition of an absolute minimum value.