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) \( \quad(x, y)=(\square \) \( \begin{array}{l}\text { minima (larger } x \text {-value) } \quad(x, y)=(\square\end{array} \)

To find the absolute extrema of the function \( f(x) = x^4 - 2x^3 + x + 1 \) on the interval \([-1, 3]\), you first need to find the critical points in that interval by calculating the derivative \( f'(x) \) and setting it to zero. The derivative is: \[ f'(x) = 4x^3 - 6x^2 + 1 \] Setting \( f'(x) = 0 \): \[ 4x^3 - 6x^2 + 1 = 0 \] This cubic equation can be solved using a graphing utility or numerical methods to find critical points. Once you have the critical points within the interval \([-1, 3]\), evaluate \( f(x) \) at these points as well as at the endpoints \( x = -1 \) and \( x = 3 \). After calculating \( f(-1) \), \( f(3) \), and \( f\) at your critical points, you'll compare these values to determine the absolute minimum and maximum on the interval. If calculated correctly, you would find: - For minima: - Smaller \( (x, y) \): \( (-1, 3) \) - Larger \( (x, y) \): \( (1, 1) \) Thus: minima (smaller \( x \)-value) \( (x, y) = (-1, 3) \) minima (larger \( x \)-value) \( (x, y) = (1, 1) \)