Post a New Question

calculus

posted by .

Use analytic methods to find the global extreme values of the function on the interval and state where they occur.

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

  • calculus -

    Find 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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question