Determine the maximum and minimum of each function on the given interval.
a) ���� = 2x^3�� − 9x^2�� ,−2 ≤ x� ≤ 4
b) ���� = 12x� − x^3�� , � x∈ �[−3,5]
To determine the maximum and minimum of each function on the given interval, we need to follow these steps:
a) For the function ���� = 2x^3�� − 9x^2�, where -2 ≤ x ≤ 4:
1. Calculate the derivative of the function, ����(x), by differentiating each term with respect to x.
The derivative of 2x^3 is 6x^2.
The derivative of -9x^2 is -18x.
2. Set the derivative equal to zero and solve for x to find critical points.
6x^2 - 18x = 0.
Factor out x: x(6x - 18) = 0.
Set each factor equal to zero: x = 0 or 6x - 18 = 0.
Solve for x: x = 0 or x = 3.
3. Identify the endpoints of the interval, -2 and 4.
4. Evaluate the original function at the critical points and endpoints.
For x = -2: ����(-2) = 2(-2)^3 - 9(-2)^2 = -4 + 36 = 32.
For x = 0: ����(0) = 2(0)^3 - 9(0)^2 = 0.
For x = 3: ����(3) = 2(3)^3 - 9(3)^2 = 54 - 81 = -27.
For x = 4: ����(4) = 2(4)^3 - 9(4)^2 = 128 - 144 = -16.
5. Compare the function values to determine the maximum and minimum.
The maximum value is 32, which occurs at x = -2.
The minimum value is -27, which occurs at x = 3.
b) For the function ���� = 12x - x^3, where x ∈ [-3, 5]:
1. Calculate the derivative of the function, ����(x).
The derivative of 12x is 12.
The derivative of -x^3 is -3x^2.
2. Set the derivative equal to zero and solve for x to find critical points.
12 - 3x^2 = 0.
Divide by 3: 4 - x^2 = 0.
Rearrange the equation: x^2 = 4.
Take the square root: x = ±2.
The critical points are x = -2 and x = 2.
3. Identify the endpoints of the interval, -3 and 5.
4. Evaluate the original function at the critical points and endpoints.
For x = -3: ����(-3) = 12(-3) - (-3)^3 = -36 - (-27) = -9.
For x = -2: ����(-2) = 12(-2) - (-2)^3 = -24 - (-8) = -16.
For x = 2: ����(2) = 12(2) - (2)^3 = 24 - 8 = 16.
For x = 5: ����(5) = 12(5) - (5)^3 = 60 - 125 = -65.
5. Compare the function values to determine the maximum and minimum.
The maximum value is 16, which occurs at x = 2.
The minimum value is -65, which occurs at x = 5.