Questions: Find the global maximum and minimum points of the function y=3x^5-5x^3-2 on the interval [-1,3]

Solution Steps

To find the global maximum and minimum of the function \( y = 3x^5 - 5x^3 - 2 \) on the interval \([-1, 3]\), we need to follow these steps:

1. Find the derivative of the function to determine the critical points. Critical points occur where the derivative is zero or undefined.
2. Evaluate the derivative to find the critical points within the interval \([-1, 3]\).
3. Evaluate the function at the critical points and at the endpoints of the interval.
4. Compare the values obtained in the previous step to determine the global maximum and minimum.

To find the critical points of the function \( y = 3x^5 - 5x^3 - 2 \), we first calculate its derivative: \[ \frac{dy}{dx} = 15x^4 - 15x^2 \]

Set the derivative equal to zero to find the critical points: \[ 15x^4 - 15x^2 = 0 \] Factor the equation: \[ 15x^2(x^2 - 1) = 0 \] This gives the solutions: \[ x = 0, \quad x = \pm 1 \] Within the interval \([-1, 3]\), the critical points are \(x = -1, 0, 1\).

Evaluate the original function \( y = 3x^5 - 5x^3 - 2 \) at the critical points and the endpoints of the interval \([-1, 3]\):

• At \(x = -1\): \[ y(-1) = 3(-1)^5 - 5(-1)^3 - 2 = 0 \]
• At \(x = 0\): \[ y(0) = 3(0)^5 - 5(0)^3 - 2 = -2 \]
• At \(x = 1\): \[ y(1) = 3(1)^5 - 5(1)^3 - 2 = -4 \]
• At \(x = 3\): \[ y(3) = 3(3)^5 - 5(3)^3 - 2 = 592 \]

Compare the values obtained:

• \(y(-1) = 0\)
• \(y(0) = -2\)
• \(y(1) = -4\)
• \(y(3) = 592\)

The global maximum value is 592 at \(x = 3\), and the global minimum value is -4 at \(x = 1\).

Final Answer