a. find the critical points of f on the given interval. b. determine the absolute extreme values of f on the…

Explanation:

Step1: Find the derivative of f(x)

Use the quotient - rule. If $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 5x$, $u'=5$, $v=(x - 8)^{\frac{1}{2}}$, $v'=\frac{1}{2}(x - 8)^{-\frac{1}{2}}$. [ \begin{align*} f'(x)&=\frac{5\sqrt{x - 8}-5x\cdot\frac{1}{2\sqrt{x - 8}}}{x - 8}\ &=\frac{\frac{10(x - 8)-5x}{2\sqrt{x - 8}}}{x - 8}\ &=\frac{10x-80 - 5x}{2(x - 8)^{\frac{3}{2}}}\ &=\frac{5x-80}{2(x - 8)^{\frac{3}{2}}} \end{align*} ]

Step2: Find the critical points

Set $f'(x)=0$ and solve for $x$. [ \begin{align*} \frac{5x - 80}{2(x - 8)^{\frac{3}{2}}}&=0\ 5x-80&=0\ x& = 16 \end{align*} ] Also, check where $f'(x)$ is undefined. The derivative $f'(x)$ is undefined when $x = 8$, but $8\notin[10,40]$. So the only critical point in the interval $[10,40]$ is $x = 16$.

Step3: Evaluate f(x) at critical and endpoints

Evaluate $f(x)$ at $x = 10$, $x=16$ and $x = 40$. [ \begin{align*} f(10)&=\frac{5\times10}{\sqrt{10 - 8}}=\frac{50}{\sqrt{2}} = 25\sqrt{2}\approx35.36\ f(16)&=\frac{5\times16}{\sqrt{16 - 8}}=\frac{80}{\sqrt{8}}=\frac{80}{2\sqrt{2}}=20\sqrt{2}\approx28.28\ f(40)&=\frac{5\times40}{\sqrt{40 - 8}}=\frac{200}{\sqrt{32}}=\frac{200}{4\sqrt{2}}=25\sqrt{2}\approx35.36 \end{align*} ]

Step4: Determine absolute extreme values

The absolute minimum value of $f(x)$ on $[10,40]$ is $20\sqrt{2}$ at $x = 16$, and the absolute maximum value is $25\sqrt{2}$ at $x=10$ and $x = 40$.

Answer:

a. The critical point in the interval $[10,40]$ is $x = 16$. b. The absolute minimum value is $20\sqrt{2}$ at $x = 16$, and the absolute maximum value is $25\sqrt{2}$ at $x=10$ and $x = 40$. c. (This part is for using a graphing utility which is not shown here. But by graphing $y=\frac{5x}{\sqrt{x - 8}}$ on $[10,40]$, one can confirm the above - found extreme values and critical point.)