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May 29, 2016
Posted by **Jocelyn** on Saturday, February 26, 2011 at 6:24pm.

y=x(2-x)^(1/2), -2<=x<=2

- calculus -
**agrin04**, Sunday, February 27, 2011 at 6:35amFind the first derivative of the function:

dy/dx = (2-x)^(1/2) - (1/2)x.(2-x)^(-1/2)

= (4-3x)/2sqrt(2-x)

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 non-imaginary 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