Use a graphing utility to graph the function and find the absolute extrema of the function on the given Interval. (Round your answers to three decimal places. If an answer does not exist, enter DNE.) $$ f(x)=x^{4}-2 x^{3}+x+1, \quad[-1,3] $$ minima (smaller $$ x $$-value) $$ \quad(x, y)=(\square) $$ minima (larger $$ x $$-value) $$ \quad(x, y)=(\square) $$ maximum $$ (x, y)=(\square) $$

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To find the absolute extrema of the function $$ f(x) = x^{4} - 2x^{3} + x + 1 $$ on the interval $$[-1, 3]$$, we need to find the minimum and maximum values of the function within this interval.

First, let's find the critical points of the function by taking the derivative and setting it equal to zero:

$$ f'(x) = 4x^{3} - 6x^{2} + 1 $$

Setting $$ f'(x) = 0 $$, we get:

$$ 4x^{3} - 6x^{2} + 1 = 0 $$

Solving this equation, we find the critical points:

$$ x = 1 $$

Now, we need to evaluate the function at the critical points and at the endpoints of the interval to find the absolute extrema.

Let's calculate the function values at the critical points and the endpoints:

- At $$ x = -1 $$:
$$ f(-1) = (-1)^{4} - 2(-1)^{3} + (-1) + 1 = 1 + 2 - 1 + 1 = 3 $$

- At $$ x = 1 $$:
$$ f(1) = (1)^{4} - 2(1)^{3} + (1) + 1 = 1 - 2 + 1 + 1 = 1 $$

- At $$ x = 3 $$:
$$ f(3) = (3)^{4} - 2(3)^{3} + (3) + 1 = 81 - 54 + 3 + 1 = 31 $$

Now, we can determine the absolute extrema:

- Minimum (smaller $$ x $$-value): $$ (x, y) = (-1, 3) $$
- Minimum (larger $$ x $$-value): $$ (x, y) = (1, 1) $$
- Maximum: $$ (x, y) = (3, 31) $$
Absolute extrema of $$ f(x) = x^{4} - 2x^{3} + x + 1 $$ on $$[-1, 3]$$: - Minimum (smaller $$ x $$-value): $$ (-1, 3) $$ - Minimum (larger $$ x $$-value): $$ (1, 1) $$ - Maximum: $$ (3, 31) $$

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