determine all inflection points of f(x)=(18x + 9)(x - 5)^(2/3)+3x - 1.as your answer, please enter the sum…

Explanation:

Step1: Find the first - derivative

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 18x + 9$ and $v=(x - 5)^{2/3}$, and the sum rule for differentiation. The derivative of $u=18x + 9$ is $u^\prime=18$, and the derivative of $v=(x - 5)^{2/3}$ is $v^\prime=\frac{2}{3}(x - 5)^{-1/3}$. The derivative of $3x-1$ is $3$. So, $f^\prime(x)=18(x - 5)^{2/3}+(18x + 9)\frac{2}{3}(x - 5)^{-1/3}+3$. Simplify $f^\prime(x)$: [ \begin{align*} f^\prime(x)&=18(x - 5)^{2/3}+\frac{2(18x + 9)}{3(x - 5)^{1/3}}+3\ &=\frac{18(x - 5)+2(18x + 9)+3\times3(x - 5)^{1/3}}{(x - 5)^{1/3}}\ &=\frac{18x-90 + 36x+18 + 9(x - 5)^{1/3}}{(x - 5)^{1/3}}\ &=\frac{54x-72 + 9(x - 5)^{1/3}}{(x - 5)^{1/3}} \end{align*} ]

Step2: Find the second - derivative

Use the quotient rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = 54x-72 + 9(x - 5)^{1/3}$ and $v=(x - 5)^{1/3}$. $u^\prime=54 + 9\times\frac{1}{3}(x - 5)^{-2/3}=54 + 3(x - 5)^{-2/3}$ and $v^\prime=\frac{1}{3}(x - 5)^{-2/3}$. [ \begin{align*} f^{\prime\prime}(x)&=\frac{(54 + 3(x - 5)^{-2/3})(x - 5)^{1/3}-(54x-72 + 9(x - 5)^{1/3})\frac{1}{3}(x - 5)^{-2/3}}{(x - 5)^{2/3}}\ &=\frac{54(x - 5)^{1/3}+3(x - 5)^{-1/3}-\frac{54x-72}{3}(x - 5)^{-2/3}- 3(x - 5)^{-1/3}}{(x - 5)^{2/3}}\ &=\frac{54(x - 5)^{1/3}-\frac{54x-72}{3}(x - 5)^{-2/3}}{(x - 5)^{2/3}}\ &=\frac{54(x - 5)- (54x-72)}{3(x - 5)^{4/3}}\ &=\frac{54x-270-54x + 72}{3(x - 5)^{4/3}}\ &=\frac{-198}{3(x - 5)^{4/3}}=\frac{-66}{(x - 5)^{4/3}} \end{align*} ] The second - derivative $f^{\prime\prime}(x)$ is never equal to zero since the numerator is a non - zero constant ($-66$). But $f^{\prime\prime}(x)$ is undefined at $x = 5$. We check the concavity on either side of $x = 5$. For $x\neq5$, $f^{\prime\prime}(x)<0$. So, the function $y = f(x)$ is concave down on $(-\infty,5)\cup(5,\infty)$. There are no inflection points.

Answer:

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