\begin{tabular}{l} (1 point) Consider the function \( f(x)=4-2 x^{2},-4 \leq x \leq 1 \) \\ The absolute maximum value is \\ and this occurs at \( x \) equals \\ The absolute minimum value is \\ and this occurs at \( x \) equals \\ \hline\end{tabular}

To find the absolute maximum and minimum values of the function \( f(x) = 4 - 2x^2 \) on the interval \( -4 \leq x \leq 1 \), 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 the endpoints of the interval. 3. **Determine the absolute maximum and minimum values** from these evaluations. ### Step 1: Find the critical points First, we calculate the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx}(4 - 2x^2) = -4x \] Next, we set the derivative equal to zero to find the critical points: \[ -4x = 0 \implies x = 0 \] ### Step 2: Evaluate the function at critical points and endpoints Now we will evaluate \( f(x) \) at the critical point \( x = 0 \) and at the endpoints \( x = -4 \) and \( x = 1 \). 1. **At \( x = -4 \)**: \[ f(-4) = 4 - 2(-4)^2 = 4 - 2(16) = 4 - 32 = -28 \] 2. **At \( x = 0 \)**: \[ f(0) = 4 - 2(0)^2 = 4 - 0 = 4 \] 3. **At \( x = 1 \)**: \[ f(1) = 4 - 2(1)^2 = 4 - 2 = 2 \] ### Step 3: Determine the absolute maximum and minimum values Now we compare the values obtained: - \( f(-4) = -28 \) - \( f(0) = 4 \) - \( f(1) = 2 \) From these evaluations, we can conclude: - The **absolute maximum value** is \( 4 \) and it occurs at \( x = 0 \). - The **absolute minimum value** is \( -28 \) and it occurs at \( x = -4 \). ### Final Answer The absolute maximum value is \( 4 \) and this occurs at \( x \) equals \( 0 \). The absolute minimum value is \( -28 \) and this occurs at \( x \) equals \( -4 \).