a graphing device is recommended. a function is given. u(x)=x√(5 - x) (a) find the local maximum value of…

Explanation:

Step1: Find the derivative of $U(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x$ and $v=\sqrt{5 - x}=(5 - x)^{\frac{1}{2}}$. $u^\prime=1$, $v^\prime=\frac{1}{2}(5 - x)^{-\frac{1}{2}}\times(- 1)=-\frac{1}{2\sqrt{5 - x}}$. $U^\prime(x)=\sqrt{5 - x}-\frac{x}{2\sqrt{5 - x}}=\frac{2(5 - x)-x}{2\sqrt{5 - x}}=\frac{10 - 2x - x}{2\sqrt{5 - x}}=\frac{10 - 3x}{2\sqrt{5 - x}}$.

Step2: Find the critical points

Set $U^\prime(x)=0$, then $\frac{10 - 3x}{2\sqrt{5 - x}} = 0$. The numerator must be zero, so $10 - 3x=0$, which gives $x=\frac{10}{3}\approx3.33$. Also, the function $U(x)$ is defined for $x\leqslant5$.

Step3: Determine the intervals of increase and decrease

Choose test - points in the intervals $(-\infty,\frac{10}{3})$ and $(\frac{10}{3},5)$. For the interval $[0,\frac{10}{3})$, let's choose $x = 1$. Then $U^\prime(1)=\frac{10-3\times1}{2\sqrt{5 - 1}}=\frac{7}{4}>0$, so the function is increasing on the interval $[0,3.33]$. For the interval $(\frac{10}{3},5]$, let's choose $x = 4$. Then $U^\prime(4)=\frac{10-3\times4}{2\sqrt{5 - 4}}=\frac{10 - 12}{2}=-1<0$, so the function is decreasing on the interval $[3.33,5]$.

Answer:

increasing: $[0,3.33]$ decreasing: $[3.33,5]$