PREVIOUS ANSWERS 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.) \[ f(x)=x^{4}-2 x^{3}+x+1,[-1,3] \] minima (smaller \( x \)-value) \( \quad(x, y)=(\square) \)

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 use a graphing utility to visualize the function within the given interval. This will help me identify any local maxima or minima. 2. **Find Critical Points**: To find the absolute extrema, I need to determine the critical points of the function. Critical points occur where the derivative is zero or undefined. So, I'll compute the first derivative of \( f(x) \). \[ f'(x) = \frac{d}{dx}(x^4 - 2x^3 + x + 1) = 4x^3 - 6x^2 + 1 \] 3. **Set the Derivative Equal to Zero**: To find the critical points, I'll set \( f'(x) = 0 \) and solve for \( x \). \[ 4x^3 - 6x^2 + 1 = 0 \] This is a cubic equation, which can be challenging to solve analytically. I'll use numerical methods or a graphing calculator to approximate the roots. 4. **Approximate the Roots**: Using a graphing utility, I'll plot \( f'(x) = 4x^3 - 6x^2 + 1 \) to estimate the points where it crosses the x-axis. - The graph shows that there are two real roots within the interval \([-1, 3]\). - Let's approximate these roots: - \( x \approx 0.25 \) - \( x \approx 1.5 \) 5. **Evaluate the Function at Critical Points and Endpoints**: To find the absolute extrema, I'll evaluate \( f(x) \) at the critical points and the endpoints of the interval. - **At \( x = -1 \):** \[ f(-1) = (-1)^4 - 2(-1)^3 + (-1) + 1 = 1 + 2 - 1 + 1 = 3 \] - **At \( x \approx 0.25 \):** \[ f(0.25) = (0.25)^4 - 2(0.25)^3 + 0.25 + 1 \approx 0.0039 - 0.0156 + 0.25 + 1 \approx 1.2383 \] - **At \( x \approx 1.5 \):** \[ f(1.5) = (1.5)^4 - 2(1.5)^3 + 1.5 + 1 \approx 5.0625 - 6.75 + 1.5 + 1 \approx 0.8125 \] - **At \( x = 3 \):** \[ f(3) = 3^4 - 2(3)^3 + 3 + 1 = 81 - 54 + 3 + 1 = 31 \] 6. **Determine Absolute Extrema**: - **Absolute Maximum**: The highest value among these is \( f(3) = 31 \). - **Absolute Minimum**: The lowest value is \( f(1.5) \approx 0.8125 \). 7. **Conclusion**: The absolute maximum of \( f(x) \) on the interval \([-1, 3]\) is 31 at \( x = 3 \), and the absolute minimum is approximately 0.8125 at \( x \approx 1.5 \).