17. 0.48/0.71 points details my notes previous answers ask your teacher sprecalc8 2.3.059. practice another…

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) and (v^\prime=\frac{1}{2}(5 - x)^{-\frac{1}{2}}\times(- 1)=-\frac{1}{2\sqrt{5 - x}}). Then (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 (since the denominator cannot be zero for the function to be well - defined), so (10 - 3x=0), which gives (x=\frac{10}{3}\approx3.33). Also, the function (U(x)) is defined for (x\leqslant5) (because of the square - root (\sqrt{5 - x})).

Step3: Determine the intervals of increase and decrease

Choose test points in the intervals ((-\infty, \frac{10}{3})) and ((\frac{10}{3},5)). Let's choose (x = 0) for the first interval. (U^\prime(0)=\frac{10-3\times0}{2\sqrt{5 - 0}}=\frac{10}{2\sqrt{5}}>0), so the function is increasing on the interval ([0,\frac{10}{3}]). Let's choose (x = 4) for the second interval. (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 ([\frac{10}{3},5]).

Answer:

Increasing interval: ([0,3.33]) Decreasing interval: ([3.33,5])