To find the absolute maximum and minimum values of the function $f(x) = 4 - 2x^2$ over the interval $-4 \leq x \leq 1$, let's evaluate the function's behavior within the interval, checking both endpoints and any critical points in this range.

1. Find critical points by taking the derivative and setting it equal to zero:

   $f(x) = 4 - 2x^2$ $f'(x) = -4x$

   Set $f'(x) = 0$: $-4x = 0$ $x = 0$

   The critical point is $x = 0$, which is within the interval $[-4, 1]$.

2. Evaluate the function at the endpoints and critical point:

   • At $x = -4$: $f(-4) = 4 - 2(-4)^2 = 4 - 2 \cdot 16 = 4 - 32 = -28$

   • At $x = 1$: $f(1) = 4 - 2(1)^2 = 4 - 2 = 2$

   • At $x = 0$ (critical point): $f(0) = 4 - 2(0)^2 = 4$

3. Determine the absolute maximum and minimum:

   • The absolute maximum value of $f(x)$ is $4$, which occurs at $x = 0$.
   • The absolute minimum value of $f(x)$ is $-28$, which occurs at $x = -4$.

Summary

• Absolute Maximum Value: $4$, occurring at $x = 0$
• Absolute Minimum Value: $-28$, occurring at $x = -4$

Would you like further details on any part of the process?

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1. How do we find critical points in other types of functions?
2. What is the importance of checking endpoints in closed intervals?
3. Can absolute extrema occur at points other than endpoints or critical points?
4. How do we interpret the derivative of a function geometrically?
5. What is the impact of a negative sign in front of an $x^2$ term in functions?

Tip: Always evaluate both endpoints and critical points when determining absolute extrema on a closed interval.