Expert: Scotto - 11/3/2008

Question
QUESTION: I was given the problem y=x times the square root of (3-x) and I need to find the relative max and min.  I get that the first derivative is (3-x)^(1/2)-((x)/2(3-x)^(-1/2)), but I don't know how to go from there.  PLEASE HELP!!

ANSWER: Is the problem x√(3-x)?
The derivative is √(3-x) - x/(2√(3-x)).
I believe that's what I read.
The derivative of

Setting this to 0 gives √(3-x) - x/(2√(3-x)) = 0.
Multiply by 2√(3-x) and you get 2(3-x) - x = 0 (note that this means that x can't be 3).
Multiply and get 6 - 2x - x = 0, or 6 - 3x = 0.
This says that there is a point at x=2 that is either a min or a max.

If the original function is looked at, it can be seen to be 0 at x=0 and x=3.  Inbetween these two points it is positive, so that at x=2, there must be a maximum.  When x is greater than 3, the function is undefined.  Knowing this, there must be a local minimum at x=3.

To graph the function, note that the function is 0 at x=0 and x=3.  At x=2, the function is 2.  The function is concave down and not defined for x>3.

---------- FOLLOW-UP ----------

QUESTION: So using the second derivative, how would you find the points of inflection? Thanks so much for answering my questions.

Answer
The inflection points are the points that are between where the curve is concave up and concave down.  Since when the second derivative is positive means the function is concave up and when it is negative means it is concave down, then where it is 0 is an inflection point.

Thank you for asking these wonderful questions to probe my mind for answers.