Since f(x)=√(x-3) has an absolute minimum of 0 does its critical point exist?

Question

I'm really confused because someone mentioned that f(x)=√(x+3) does not have a critical point.
Does that mean that if there is an end point there, there can't be a critical point?

Answer 1

f(x) has an absolute minimum of 0 because of the definition of the square root.

Since we cannot take the square root of a negative number,
x-3 ≥ 0
x ≥ 3
and f(3) = 0

and f(x) only starts at x = 3

If you sketch y = √(x-3) , there are no points below the x-axis nor to the left of (3,0) and (3,0) is the minimum point.

It depends how your text or your instructor has defined "critical" point.

Answer 2

I'd say the critical point exists, since the usual definition that y'=0 or is undefined.

In either case, I think that (3,0) is the critical point.

Answer 3

Would -3,0 be the critical point for f(x)=√(x+3)?