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
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
To find the global extreme values of the function y = x(2-x)^(1/2) on the interval [-2, 2], we need to follow these steps:
1. Find the critical points of the function within the given interval.
2. Check the values of the function at the critical points and the endpoints of the interval.
3. Determine which values correspond to the global maximum and minimum.
Let's begin by finding the critical points of the function:
Step 1: Find the derivative of the function.
Taking the derivative of y = x(2-x)^(1/2) with respect to x will help us locate the critical points.
Let's use the product rule to differentiate the function:
dy/dx = [d(x)/dx][(2-x)^(1/2)] + x [d(2 - x)^(1/2)/dx]
dy/dx = [(2-x)^(1/2)] + x * 1/2 * (2 - x)^(-1/2) * (-1)
Simplifying this expression, we get:
dy/dx = (2-x)^(1/2) - x/(2(2-x)^(1/2))
Step 2: Solve for dy/dx = 0 to find the critical points.
Setting dy/dx = 0 and solving for x, we get:
(2-x)^(1/2) - x/(2(2-x)^(1/2)) = 0
Multiplying both sides of the equation by 2(2-x)^(1/2), we have:
(2-x) - x^2 = 0
Expanding and rearranging terms:
x^2 - 2x + 2 = 0
Using the quadratic formula, we find the solutions to be:
x = (2 ± √(-4))/2
= 1 ± i√3
Since the solutions are complex numbers, there are no critical points within the given interval [-2, 2].
Now, let's move on to Step 2:
Step 3: Check the values of the function at the endpoints and verify if they correspond to global maximum or minimum.
a) Evaluate the function at x = -2:
y(-2) = -2(2-(-2))^(1/2)
= -2(2+2)^(1/2)
= -2(4)^(1/2)
= -2(2)
= -4
b) Evaluate the function at x = 2:
y(2) = 2(2-2)^(1/2)
= 2(0)^(1/2)
= 2(0)
= 0
Since we only have one critical point that lies outside the given interval (-2 <= x <= 2) and no other critical points, we only need to compare the values at the endpoints.
Therefore, the global maximum value occurs at x = -2, with y = -4, and the global minimum value occurs at x = 2, with y = 0.
In summary, the global maximum value of the function y = x(2-x)^(1/2) occurs at x = -2, with y = -4, and the global minimum value occurs at x = 2, with y = 0.
To find the global extreme values of the function y = x(2 - x)^(1/2) on the interval -2 ≤ x ≤ 2, we can follow these steps:
1. Find the critical points of the function by setting the derivative equal to zero.
2. Determine the endpoints of the interval.
3. Evaluate the function at the critical points and endpoints.
4. Compare the values obtained in step 3 to identify the global maximum and minimum.
Let's begin:
1. Find the derivative of the function with respect to x:
y' = (2 - x)^(1/2) - (x / (2 - x)^(1/2))
Setting y' = 0:
(2 - x)^(1/2) - (x / (2 - x)^(1/2)) = 0
Rearranging this equation, we get:
(2 - x)^(1/2) = (x / (2 - x)^(1/2))
Squaring both sides:
2 - x = x^2 / (2 - x)
Multiplying both sides by (2 - x):
2(2 - x) - x(2 - x) = x^2
Simplifying the equation:
4 - 2x - 2x + x^2 = x^2
4 - 4x = 0
4x = 4
x = 1
The critical point is x = 1.
2. Next, we need to determine the values of y at the endpoints of the interval -2 ≤ x ≤ 2.
Evaluating the function at x = -2:
y = (-2)(2 - (-2))^(1/2)
= (-2)(2 + 2)^(1/2)
= (-2)(4)^(1/2)
= (-2)(2)
= -4
Evaluating the function at x = 2:
y = (2)(2 - 2)^(1/2)
= (2)(0)^(1/2)
= (2)(0)
= 0
The endpoints give us y = -4 and y = 0.
3. Evaluating the function at the critical point:
y = (1)(2 - 1)^(1/2)
= (1)(1)^(1/2)
= 1
The critical point gives us y = 1.
4. Comparing the values, we have:
y = -4, y = 0 (endpoints)
y = 1 (critical point)
The global maximum value is 1, which occurs at x = 1.
The global minimum value is -4, which occurs at x = -2.