Questions: Find the global maximum and global minimum of f(x)=x^3-3x+2 on the following intervals. a) [-3,3] b) [-5,1] c) [-1.5,1]

Transcript text: Find the global maximum and global minimum of $f(x)=x^{3}-3 x+2$ on the following intervals. a) $[-3,3]$ b) $[-5,1]$ c) $[-1.5,1]$

Solution

Solution Steps

To find the global maximum and minimum of a function on a closed interval, we need to evaluate the function at critical points and endpoints of the interval. First, find the derivative of the function and solve for critical points by setting the derivative to zero. Then, evaluate the function at these critical points and at the endpoints of the interval. Compare these values to determine the global maximum and minimum.

Step 1: Find the Critical Points

The function is given by $ f(x) = x^3 - 3x + 2 $. To find the critical points, we first compute the derivative: \[ f'(x) = 3x^2 - 3 \] Setting the derivative equal to zero gives: \[ 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = -1, 1 \] Thus, the critical points are $ x = -1 $ and $ x = 1 $.

Step 2: Evaluate the Function at the Endpoints and Critical Points

We will evaluate $ f(x) $ at the critical points and the endpoints of each interval.

a) Interval $[-3, 3]$

Endpoints: $ f(-3) = (-3)^3 - 3(-3) + 2 = -27 + 9 + 2 = -16 $
Critical points:
- $ f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4 $
- $ f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 $

Comparing these values: \[ \text{Max} = 4, \quad \text{Min} = -16 \]

b) Interval $[-5, 1]$

Endpoints: $ f(-5) = (-5)^3 - 3(-5) + 2 = -125 + 15 + 2 = -108 $
Critical points:
- $ f(-1) = 4 $ (as calculated previously)
- $ f(1) = 0 $ (as calculated previously)

Comparing these values: \[ \text{Max} = 4, \quad \text{Min} = -108 \]

c) Interval $[-1.5, 1]$

Endpoints: $ f(-1.5) = (-1.5)^3 - 3(-1.5) + 2 = -3.375 + 4.5 + 2 = 3.125 $
Critical points:
- $ f(-1) = 4 $ (as calculated previously)
- $ f(1) = 0 $ (as calculated previously)

Comparing these values: \[ \text{Max} = 4, \quad \text{Min} = 0 \]