for f(x)=(6x + 15)∛((x - 15)²)+4, use the second - derivative test to determine the local maximum of f. as…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 6x + 15$ and $v=(x - 15)^{\frac{2}{3}}$. $u^\prime=6$ and $v^\prime=\frac{2}{3}(x - 15)^{-\frac{1}{3}}$. $f^\prime(x)=6(x - 15)^{\frac{2}{3}}+(6x + 15)\times\frac{2}{3}(x - 15)^{-\frac{1}{3}}$ $=6(x - 15)^{\frac{2}{3}}+\frac{2(6x + 15)}{3(x - 15)^{\frac{1}{3}}}$ $=\frac{18(x - 15)+2(6x + 15)}{3(x - 15)^{\frac{1}{3}}}$ $=\frac{18x-270 + 12x+30}{3(x - 15)^{\frac{1}{3}}}$ $=\frac{30x-240}{3(x - 15)^{\frac{1}{3}}}=\frac{10x - 80}{(x - 15)^{\frac{1}{3}}}$

Step2: Find the critical points

Set $f^\prime(x)=0$, then $10x - 80 = 0$ gives $x = 8$. Also, $f^\prime(x)$ is undefined at $x = 15$.

Step3: Find the second - derivative

Use the quotient rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}$ where $u = 10x - 80$ and $v=(x - 15)^{\frac{1}{3}}$. $u^\prime = 10$ and $v^\prime=\frac{1}{3}(x - 15)^{-\frac{2}{3}}$. $f^{\prime\prime}(x)=\frac{10(x - 15)^{\frac{1}{3}}-(10x - 80)\times\frac{1}{3}(x - 15)^{-\frac{2}{3}}}{(x - 15)^{\frac{2}{3}}}$ $=\frac{30(x - 15)-(10x - 80)}{3(x - 15)^{\frac{4}{3}}}$ $=\frac{30x-450-10x + 80}{3(x - 15)^{\frac{4}{3}}}$ $=\frac{20x-370}{3(x - 15)^{\frac{4}{3}}}$

Step4: Evaluate the second - derivative at critical points

For $x = 8$: $f^{\prime\prime}(8)=\frac{20\times8-370}{3(8 - 15)^{\frac{4}{3}}}=\frac{160-370}{3\times(- 7)^{\frac{4}{3}}}=\frac{-210}{3\times49^{\frac{1}{3}}}<0$. So $x = 8$ is a local maximum. For $x = 15$, $f^{\prime\prime}(x)$ is undefined, so we can't use the second - derivative test at $x = 15$.

Step5: Find the value of $f$ at the local maximum

$f(8)=(6\times8 + 15)\sqrt[3]{(8 - 15)^{2}}+4=(48 + 15)\sqrt[3]{49}+4=63\sqrt[3]{49}+4$ $63\times3.6593+4=230.536+4=234.536$

Answer:

$234.536$