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 occu intervals for which the function is increasing or decreasing. Round to the nearest hundredth, if necessary. From least to greatest, the \( x \)-intercepts are \( x \approx \square, x=1 \), and \( x \approx \square \) The local maximum is \( \square, \square \). The local minimum is \( \square, \square \). The function is increasing when \( x<\square \) and \( x>\square \) The function is decreasing when \( \square

Ask by Bowen Rodriquez. in the United States

Feb 04,2025

To find the \( x \)-intercepts, we need to solve the equation \( f(x) = 0 \) for \( x \). Given function: \( f(x) = x^{5} - 4x^{3} + x^{2} + 2 \) To find the \( x \)-intercepts, we set \( f(x) = 0 \) and solve for \( x \): \( x^{5} - 4x^{3} + x^{2} + 2 = 0 \) To find the local maximums and local minimums, we need to find the critical points by taking the derivative of the function and setting it equal to zero. The derivative of \( f(x) \) is: \( f'(x) = 5x^{4} - 12x^{2} + 2x \) Setting \( f'(x) = 0 \) to find the critical points: \( 5x^{4} - 12x^{2} + 2x = 0 \) Solving this equation will give us the critical points where the local maximums and local minimums occur. Let's solve the equation \( 5x^{4} - 12x^{2} + 2x = 0 \) to find the critical points. Solve the equation by following steps: - step0: Solve for \(x\): \(5x^{4}-12x^{2}+2x=0\) - step1: Factor the expression: \(x\left(5x^{3}-12x+2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x=0\\&5x^{3}-12x+2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=0\\&x\approx -1.626625\\&x\approx 0.168666\\&x\approx 1.457959\end{align}\) - step4: Rewrite: \(x_{1}\approx -1.626625,x_{2}=0,x_{3}\approx 0.168666,x_{4}\approx 1.457959\) The critical points are approximately \( x \approx -1.626625, x = 0, x \approx 0.168666, x \approx 1.457959 \). Now, let's find the local maximums and local minimums by analyzing the critical points. 1. Local Maximums: - The local maximum occurs at \( x \approx 0.168666 \). 2. Local Minimums: - The local minimum occurs at \( x \approx -1.626625 \). Now, let's determine the intervals where the function is increasing or decreasing. 1. Increasing Intervals: - The function is increasing when \( x < -1.626625 \) and \( x > 1.457959 \). 2. Decreasing Intervals: - The function is decreasing when \( -1.626625 < x < 0.168666 \). Therefore, the \( x \)-intercepts are approximately \( x \approx -1.626625, x = 1, x \approx 1.457959 \). The local maximum is approximately \( x \approx 0.168666 \). The local minimum is approximately \( x \approx -1.626625 \). The function is increasing when \( x < -1.626625 \) and \( x > 1.457959 \). The function is decreasing when \( -1.626625 < x < 0.168666 \).