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) $$ \quad(x, y)=(\square) $$

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Alright, I need to find the absolute extrema of the function $$ f(x) = x^4 - 2x^3 + x + 1 $$ on the interval $$[-1, 3]$$. To do this, I'll follow these steps:

1. **Graph the Function**: First, I'll graph the function to visualize its behavior on the given interval. This will help me identify any local maxima or minima.

2. **Find Critical Points**: To find the extrema, I'll need to find the critical points of the function. Critical points occur where the derivative is zero or undefined. So, I'll compute the first derivative of $$ f(x) $$ and set it equal to zero to solve for $$ x $$.

3. **Determine the Nature of Critical Points**: Once I have the critical points, I'll use the second derivative test to determine whether each critical point is a local maximum, local minimum, or neither.

4. **Evaluate the Function at Critical Points and Endpoints**: After identifying the critical points, I'll evaluate the function at these points and at the endpoints of the interval $$[-1, 3]$$ to find the absolute maximum and minimum values.

5. **Round the Answers**: Finally, I'll round the answers to three decimal places as required.

**Step 1: Graph the Function**

I'll use a graphing utility to plot $$ f(x) = x^4 - 2x^3 + x + 1 $$ over the interval $$[-1, 3]$$. Observing the graph will give me an initial idea of where the extrema might be located.

**Step 2: Find Critical Points**

First, I'll find the first derivative of $$ f(x) $$:
$$
f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x^3) + \frac{d}{dx}(x) + \frac{d}{dx}(1) = 4x^3 - 6x^2 + 1
$$
Setting $$ f'(x) = 0 $$:
$$
4x^3 - 6x^2 + 1 = 0
$$
This is a cubic equation, and solving it analytically might be complex. I'll use numerical methods or a graphing calculator to approximate the roots.

Using a graphing utility, I'll plot $$ f'(x) = 4x^3 - 6x^2 + 1 $$ to find the points where it crosses the x-axis. From the graph, I can estimate the critical points.

**Step 3: Determine the Nature of Critical Points**

Once I have the critical points, I'll compute the second derivative to determine their nature:
$$
f''(x) = \frac{d}{dx}(4x^3) - \frac{d}{dx}(6x^2) + \frac{d}{dx}(1) = 12x^2 - 12x
$$
For each critical point $$ x = c $$:
- If $$ f''(c) > 0 $$, it's a local minimum.
- If $$ f''(c) < 0 $$, it's a local maximum.
- If $$ f''(c) = 0 $$, the test is inconclusive.

**Step 4: Evaluate the Function at Critical Points and Endpoints**

After identifying the critical points, I'll evaluate $$ f(x) $$ at these points and at the endpoints $$ x = -1 $$ and $$ x = 3 $$ to find the absolute extrema.

**Step 5: Round the Answers**

Finally, I'll round the x and y values to three decimal places as specified.

**Potential Challenges and Considerations:**

- **Multiple Critical Points**: There might be more than one critical point, so I need to evaluate each one carefully.
- **Numerical Precision**: Rounding to three decimal places might affect the accuracy of the extrema values.
- **Graphing Accuracy**: The graphing utility's accuracy is crucial for identifying critical points and extrema accurately.

By following these steps methodically, I should be able to determine the absolute extrema of the function on the given interval.

The absolute minimum occurs at $$ (1.000, -1.000) $$.

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) \( \quad(x, y)=(\square) \)

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