To find the absolute maximum and minimum values of the function $f(x) = 2 - 6x^2$ on the interval $-4 \leq x \leq 1$, we need to evaluate $f(x)$ at the critical points within the interval and at the endpoints of the interval.

Step 1: Differentiate the function

To find the critical points, let's differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(2 - 6x^2) = -12x.$

Step 2: Set the derivative to zero

Set $f'(x) = 0$ to find the critical points: $-12x = 0 \Rightarrow x = 0.$ Since $x = 0$ lies within the interval $[-4, 1]$, it is a valid critical point.

Step 3: Evaluate $f(x)$ at the critical point and endpoints

Now we evaluate $f(x)$ at $x = -4$, $x = 1$, and $x = 0$:

1. At $x = -4$: $f(-4) = 2 - 6(-4)^2 = 2 - 6 \cdot 16 = 2 - 96 = -94.$

2. At $x = 1$: $f(1) = 2 - 6(1)^2 = 2 - 6 = -4.$

3. At $x = 0$: $f(0) = 2 - 6(0)^2 = 2.$

Step 4: Determine the absolute maximum and minimum

By comparing these values:

• The absolute maximum value is $f(0) = 2$, which occurs at $x = 0$.
• The absolute minimum value is $f(-4) = -94$, which occurs at $x = -4$.

Conclusion

• The absolute maximum value is $\mathbf{2}$, and it occurs at $x = 0$.
• The absolute minimum value is $\mathbf{-94}$, and it occurs at $x = -4$.

Would you like more details or have any questions?

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Related Questions

1. What is the impact of changing the interval on the maximum and minimum values?
2. How would the maximum and minimum values change if the function was shifted vertically?
3. What is the derivative's role in identifying maximum and minimum points?
4. How do the endpoints influence the absolute extrema of a function?
5. How does the shape of the graph of $f(x) = 2 - 6x^2$ relate to its extrema?

Tip: When finding extrema on a closed interval, always evaluate the function at both endpoints and any critical points within the interval.