For the function: \( f(x)=x^{3}-6 x^{2} \) on the interval \( [-1,5] \), The absolute min occurs at ([ ? ], [ ]) The absolute max occurs at ([ ],[ ])

To find the absolute minimum and maximum of the function \( f(x) = x^3 - 6x^2 \) on the interval \([-1, 5]\), we will follow these steps: 1. **Find the critical points** by taking the derivative of the function and setting it to zero. 2. **Evaluate the function** at the critical points and at the endpoints of the interval. 3. **Compare the values** to determine the absolute minimum and maximum. ### Step 1: Find the critical points First, we calculate the derivative of \( f(x) \): \[ f'(x) = 3x^2 - 12x \] Next, we set the derivative equal to zero to find the critical points: \[ 3x^2 - 12x = 0 \] Factoring out \( 3x \): \[ 3x(x - 4) = 0 \] This gives us the critical points: \[ x = 0 \quad \text{and} \quad x = 4 \] ### Step 2: Evaluate the function at critical points and endpoints Now we evaluate \( f(x) \) at the critical points and the endpoints of the interval \([-1, 5]\): - At \( x = -1 \): \[ f(-1) = (-1)^3 - 6(-1)^2 = -1 - 6 = -7 \] - At \( x = 0 \): \[ f(0) = 0^3 - 6(0)^2 = 0 \] - At \( x = 4 \): \[ f(4) = 4^3 - 6(4)^2 = 64 - 96 = -32 \] - At \( x = 5 \): \[ f(5) = 5^3 - 6(5)^2 = 125 - 150 = -25 \] ### Step 3: Compare the values Now we have the following values: - \( f(-1) = -7 \) - \( f(0) = 0 \) - \( f(4) = -32 \) - \( f(5) = -25 \) Now we can determine the absolute minimum and maximum: - The **absolute minimum** occurs at \( x = 4 \) with \( f(4) = -32 \). - The **absolute maximum** occurs at \( x = 0 \) with \( f(0) = 0 \). ### Final Answer The absolute min occurs at \([4, -32]\) and the absolute max occurs at \([0, 0]\).