A graphing calculator is recommended. A function is given. \( \begin{array}{l}U(x)=x \sqrt{4-x} \\ \text { (a) Find the local maximum value of the function and the value of } x \text { at which this occurs. State the answer rounded to two } \\ \text { ( } x, y)=(\square\end{array} \) (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer rounded to two decimal places. (Enter your answers using interval notation.) increasing decreasing

Function by following steps: - step0: Find the increasing or decreasing interval: \(U\left(x\right)=x\sqrt{4-x}\) - step1: Find the domain: \(U\left(x\right)=x\sqrt{4-x},x\leq 4\) - step2: Find the derivative: \(U^{\prime}\left(x\right)=\frac{8-3x}{2\sqrt{4-x}}\) - step3: Find the domain: \(U^{\prime}\left(x\right)=\frac{8-3x}{2\sqrt{4-x}},x<4\) - step4: Substitute \(U^{\prime}\left(x\right)=0:\) \(0=\frac{8-3x}{2\sqrt{4-x}}\) - step5: Swap the sides: \(\frac{8-3x}{2\sqrt{4-x}}=0\) - step6: Cross multiply: \(8-3x=2\sqrt{4-x}\times 0\) - step7: Simplify the equation: \(8-3x=0\) - step8: Move the constant to the right side: \(-3x=0-8\) - step9: Remove 0: \(-3x=-8\) - step10: Change the signs: \(3x=8\) - step11: Divide both sides: \(\frac{3x}{3}=\frac{8}{3}\) - step12: Divide the numbers: \(x=\frac{8}{3}\) - step13: Check if the solution is in the defined range: \(x=\frac{8}{3},x<4\) - step14: Find the intersection: \(x=\frac{8}{3}\) - step15: Determine the intervals: \(\begin{align}&x\leq \frac{8}{3}\\&\frac{8}{3}\leq x\leq 4\end{align}\) - step16: Choose the points: \(\begin{align}&x_{1}=2\\&x_{2}=3\end{align}\) - step17: Find the values of the derivatives: \(\begin{align}&U^{\prime}\left(2\right)\approx 0.707107\\&U^{\prime}\left(3\right)=-\frac{1}{2}\end{align}\) - step18: Calculate: \(\begin{align}&x\leq \frac{8}{3}\textrm{ is increasing interval}\\&\frac{8}{3}\leq x\leq 4\textrm{ is decreasing interval}\end{align}\) - step19: Evaluate: \(\begin{align}&\textrm{The increasing interval is}\textrm{ }x\leq \frac{8}{3}\\&\textrm{The decreasing interval is}\textrm{ }\frac{8}{3}\leq x\leq 4\end{align}\) Analyze the extrema of the function \( U(x)=x \sqrt{4-x} \) Function by following steps: - step0: Find the local extrema: \(U\left(x\right)=x\sqrt{4-x}\) - step1: Find the domain: \(U\left(x\right)=x\sqrt{4-x},x\leq 4\) - step2: Find the derivative: \(U^{\prime}\left(x\right)=\frac{8-3x}{2\sqrt{4-x}}\) - step3: Find the domain: \(U^{\prime}\left(x\right)=\frac{8-3x}{2\sqrt{4-x}},x<4\) - step4: Substitute \(U^{\prime}\left(x\right)=0:\) \(0=\frac{8-3x}{2\sqrt{4-x}}\) - step5: Swap the sides: \(\frac{8-3x}{2\sqrt{4-x}}=0\) - step6: Cross multiply: \(8-3x=2\sqrt{4-x}\times 0\) - step7: Simplify the equation: \(8-3x=0\) - step8: Move the constant to the right side: \(-3x=0-8\) - step9: Remove 0: \(-3x=-8\) - step10: Change the signs: \(3x=8\) - step11: Divide both sides: \(\frac{3x}{3}=\frac{8}{3}\) - step12: Divide the numbers: \(x=\frac{8}{3}\) - step13: Check if the solution is in the defined range: \(x=\frac{8}{3},x<4\) - step14: Find the intersection: \(x=\frac{8}{3}\) - step15: Determine the intervals: \(\begin{align}&\left(-\infty,\frac{8}{3}\right),\left(\frac{8}{3},4\right]\end{align}\) - step16: Choose the points: \(x_{1}=2,x_{2}=3\) - step17: Find the values of the derivatives: \(U^{\prime}\left(2\right)\approx 0.707107,U^{\prime}\left(3\right)=-\frac{1}{2}\) - step18: The local maximum is at \(x=\frac{8}{3}:\) \(\begin{align}&U\left(x\right)=x\sqrt{4-x},x=\frac{8}{3}\end{align}\) - step19: Calculate the function value: \(U\left(\frac{8}{3} \right)=\frac{16\sqrt{3}}{9}\) - step20: The local maximum of the function is \(\frac{16\sqrt{3}}{9}\) at \(x = \frac{8}{3}:\) \(\begin{align}&\textrm{The local maximum is }\frac{16\sqrt{3}}{9}\textrm{ at }x = \frac{8}{3}\end{align}\) Let's break down the solution step by step for the function \( U(x) = x \sqrt{4 - x} \). ### (a) Finding the Local Maximum Value 1. **Finding the Critical Points**: To find the local maximum, we first need to determine the critical points by finding the derivative of \( U(x) \) and setting it to zero. 2. **Calculating the Local Maximum**: The local maximum value of the function is given as \( \frac{16\sqrt{3}}{9} \) at \( x = \frac{8}{3} \). Thus, the answer for part (a) is: \[ (x, y) = \left(\frac{8}{3}, \frac{16\sqrt{3}}{9}\right) \] ### (b) Finding Intervals of Increase and Decrease 1. **Increasing Interval**: The function is increasing for \( x \leq \frac{8}{3} \). 2. **Decreasing Interval**: The function is decreasing for \( \frac{8}{3} \leq x \leq 4 \). Thus, the answers for part (b) are: - **Increasing**: \( (-\infty, \frac{8}{3}] \) - **Decreasing**: \( [\frac{8}{3}, 4] \) ### Summary of Answers - (a) \( (x, y) = \left(\frac{8}{3}, \frac{16\sqrt{3}}{9}\right) \) - (b) Increasing: \( (-\infty, \frac{8}{3}] \), Decreasing: \( [\frac{8}{3}, 4] \)