Graph \( f(x)=x^{5}-4 x^{3}+x^{2}+2 \). Identify the \( x \)-intercepts and the points where the local maximums and local minimums intervals for which the function is increasing or decreasing. Round to the nearest hundredth, if necessary. From least to greatest, the \( x \)-intercepts are \( x \approx \square, x=\square \), and \( x \approx \square \). The local maximum is \( \square, \square \). The local minimum is \( \square, \square \). The function is increasing when \( x<\square \) and \( x>\square \).

To analyze the function \( f(x)=x^{5}-4x^{3}+x^{2}+2 \), we first find the \( x \)-intercepts by setting \( f(x) = 0 \). Using numerical methods or a graphing tool, we can estimate the roots, yielding \( x \approx -2.21, x = 1, x \approx 1.21 \) as the \( x \)-intercepts. Next, we look for local extrema by finding the derivative \( f'(x) \) and setting it to zero. The derivative \( f'(x) = 5x^4 - 12x^2 + 2 \) gives us critical points, which can be approximated to yield the local maximum at \( x \approx -1.29, f(-1.29) \approx 4.01 \) and the local minimum at \( x \approx 1.29, f(1.29) \approx -1.19 \). To determine where the function is increasing and decreasing, we can analyze the sign of \( f'(x) \): - The function is increasing when \( x < -1.29 \) and \( x > 1.29 \). - The function is decreasing when \( -1.29 < x < 1.29 \). So, summarizing: From least to greatest, the \( x \)-intercepts are \( x \approx -2.21, x = 1, x \approx 1.21 \). The local maximum is \( (-1.29, 4.01) \). The local minimum is \( (1.29, -1.19) \). The function is increasing when \( x < -1.29 \) and \( x > 1.29 \).