Asked by Isabella
Since f(x)=√(x-3) has an absolute minimum of 0 does its critical point exist?
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?
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?
Answers
Answered by
Reiny
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.
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.
Answered by
Steve
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.
In either case, I think that (3,0) is the critical point.
Answered by
Isabella
Would -3,0 be the critical point for f(x)=√(x+3)?