Problem 1: Let \( f(x)=x^{3}-6 x^{2}+9 x+2 \) on the interval \( [0,3] \) - Find the absolute maximum and minimum values of \( f \) on \( [0,3] \)

To find the **absolute maximum** and **absolute minimum** of the function \( f(x) = x^{3} - 6x^{2} + 9x + 2 \) on the interval \([0, 3]\), follow these steps: ### 1. Find the Critical Points First, compute the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx}(x^{3} - 6x^{2} + 9x + 2) = 3x^{2} - 12x + 9 \] Set the derivative equal to zero to find the critical points: \[ 3x^{2} - 12x + 9 = 0 \Rightarrow x^{2} - 4x + 3 = 0 \] Factor the quadratic equation: \[ (x - 1)(x - 3) = 0 \Rightarrow x = 1 \text{ or } x = 3 \] **Critical Points:** \( x = 1 \) and \( x = 3 \) *Note:* Since \( x = 3 \) is an endpoint of the interval, the only critical point in the open interval \((0, 3)\) is \( x = 1 \). ### 2. Evaluate \( f(x) \) at Critical Points and Endpoints Calculate the value of \( f(x) \) at \( x = 0 \), \( x = 1 \), and \( x = 3 \): - **At \( x = 0 \):** \[ f(0) = 0^{3} - 6(0)^{2} + 9(0) + 2 = 2 \] - **At \( x = 1 \):** \[ f(1) = 1^{3} - 6(1)^{2} + 9(1) + 2 = 1 - 6 + 9 + 2 = 6 \] - **At \( x = 3 \):** \[ f(3) = 3^{3} - 6(3)^{2} + 9(3) + 2 = 27 - 54 + 27 + 2 = 2 \] ### 3. Determine the Absolute Maximum and Minimum - **Absolute Maximum:** The highest value of \( f(x) \) on \([0, 3]\) is **6**, occurring at \( x = 1 \). - **Absolute Minimum:** The lowest value of \( f(x) \) on \([0, 3]\) is **2**, occurring at both endpoints \( x = 0 \) and \( x = 3 \). ### **Final Answer** - **Absolute Maximum:** \( f(1) = 6 \) - **Absolute Minimum:** \( f(0) = 2 \) and \( f(3) = 2 \)