Posted by Jocelyn on .
Use analytic methods to find the global extreme values of the function on the interval and state where they occur.
y=x(2x)^(1/2), 2<=x<=2

calculus 
agrin04,
Find the first derivative of the function:
dy/dx = (2x)^(1/2)  (1/2)x.(2x)^(1/2)
= (43x)/2sqrt(2x)
You have to be careful in this matter, especially since you've encountered a square root form in the denominator.
* take the denominator, ignore the constant in front of the square root. We know that the original rule for square root that to have a nonimaginary result, the value under the root should be >= 0. However, since this form is in the denominator, the rule now becomes > 0. So:
2  x > 0
x < 2
* now take the nominator. Since the denominator value will always be positive, nominator value will be the 'decision maker', whether the graph function is increasing or decreasing.
4  3x = 0
x = 4/3
To find the global extreme values, just substitute this value to the original equation to find y
As for the behaviour, test one condition. Say that you want to know when the graph of this function is increasing
4  3x > 0
x < 4/3
So, we can say that the function is increasing at interval [2,4/3) or 2<=x<4/3, and decreasing at the interval (4/3,2) or 4/3<x<2