Consider the function $$ f(x)=1-2 x^{2}, \quad-5 \leq x \leq 1 $$ The absolute maximum value is and this occurs at $$ x=\square $$ The absolute minimum value is and this occurs at $$ x=\square $$

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1. **Determine the derivative to find critical points:**  
   The function is  
   $$
   f(x)=1-2x^2.
   $$  
   Its derivative is  
   $$
   f'(x)=-4x.
   $$  
   Setting the derivative equal to zero gives  
   $$
   -4x=0 \quad \Rightarrow \quad x=0.
   $$

2. **Evaluate the function at the critical point and the endpoints:**  
   - At $$ x=-5 $$:
     $$
     f(-5)=1-2(-5)^2=1-2(25)=1-50=-49.
     $$
   - At $$ x=0 $$:
     $$
     f(0)=1-2(0)^2=1.
     $$
   - At $$ x=1 $$:
     $$
     f(1)=1-2(1)^2=1-2=-1.
     $$

3. **Identify the absolute maximum and minimum:**  
   - The absolute maximum value is $$ f(0)=1 $$ and occurs at $$ x=0 $$.  
   - The absolute minimum value is $$ f(-5)=-49 $$ and occurs at $$ x=-5 $$.

**Final Answers:**  
- The absolute maximum value is $$ 1 $$ and this occurs at $$ x=0 $$.  
- The absolute minimum value is $$ -49 $$ and this occurs at $$ x=-5 $$.
The absolute maximum value is 1, occurring at $$ x=0 $$, and the absolute minimum value is -49, occurring at $$ x=-5 $$.

Consider the function
The absolute maximum value is
and this occurs at
The absolute minimum value is
and this occurs at

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Consider the function The absolute maximum value is and this occurs at The absolute minimum value is and this occurs at