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How to find the inflection points of the function

$$x(6-x)^{2/3}$$

I have tried doing it by putting $f"x=0$ and it was coming out to be a polynomial. Is there any method to check inflection points using first derivative method?

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$$f''(x)=\frac{2}{9}\frac{-36+5x}{(6-x)^{4/3}}$$ $$f''(x)=0$$ if $$x=\frac{36}{5}$$ can you proceed?

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Inflection points are defined as passing from convexity to concavity, which is equivalent as passing from an increasing $f^\prime$ to a decreasing $f^\prime$ for differentiable functions, its extremas. I am not familiar of ways of getting such information from $f^\prime$ without differentiating.

$f(x) = x(6-x)^{\frac 2 3}$

However, we can compute that $$f^\prime(x) = (6-x)^{\frac 2 3} - \frac 2 3 x(6-x)^{-\frac 1 3} = (6-x)^{-\frac 1 3}((6 - x) - \frac 2 3 x) = (6-x)^{-\frac 1 3}(6 - \frac 5 3x)$$

Therefore $$f^{\prime\prime}(x) = \frac 1 3(6-x)^{-\frac 4 3}(6-\frac 5 3x) - \frac 5 3 (6-x)^{-\frac 1 3} = \frac 1 3(6-x)^{-\frac 1 3}((6-x)^{-1}(6-\frac 5 3x) - 5)$$

We search for $f^{\prime\prime}(x) = 0 \Rightarrow x = 6$ (Cannot be, the function is not differentiable there) or $\frac {6 - \frac 5 3 x} {6 - x} = 5 \Rightarrow 6-\frac 5 3 x = 30 - 5x \Rightarrow \frac {10} 3 x = 24 \Rightarrow x = \frac {72} {10} = \frac {36} {5}$.

In conclusion: the inflection point is $x = \frac {36} {5}$

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We have $$f(x)=x(6-x)^{2/3}$$

We can compute $$f'(x)=\frac{d}{dx}x(6-x)^{2/3}=\frac{18-5x}{3\sqrt[3]{6-x}}$$

We then set this equal to $0$ to find critical points:

\begin{align}\frac{18-5x}{3\sqrt[3]{6-x}}&=0\\ 18-5x&=0\times\left(3\sqrt[3]{6-x}\right)\\ 18-5x&=0\\ 5x&=18\\ x&=\frac{18}{5}\end{align}

This claim is backed up by WolframAlpha

We then test this value against the second derivative: $$f''(x)=\frac{d^2}{dx^2}x(6-x)^{2/3}=\frac{10x-72}{9(6-x)^{4/3}}$$

The necessary condition for $x$ to be an inflection point is for $f''(x)=0$

So we test \begin{align}f''\left(\frac{18}{5}\right)&=\frac{10\times\frac{18}{5}-72}{9\left(6-\frac{18}{5}\right)^{4/3}}\\\\ &=-\frac{36}{9\left(\frac{12}{5}\right)^{4/3}}\\\\ &=-\frac{4}{\left(\frac{12}{5}\right)^{4/3}}\\\\ &=-\frac{20\sqrt[3]{\frac 53}}{12\times 2^{2/3}}\\\\ &=-\frac{5\sqrt[3]{\frac 53}}{3\times 2^{2/3}}\\\\ &\neq 0\end{align}

Therefore we can say that this function has no points of inflection

This claim is backed up by WolframAlpha

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    $\begingroup$ The points of inflection are not the points of extremum. $\endgroup$
    – AsafHaas
    May 16 '17 at 14:48
  • $\begingroup$ You answer does not address the question. $\endgroup$
    – Jean Marie
    May 16 '17 at 14:48
  • $\begingroup$ @AsafHaas My bad, I'll edit it and correct it $\endgroup$
    – lioness99a
    May 16 '17 at 14:50
  • $\begingroup$ @JeanMarie I misread it - am editing to answer the question $\endgroup$
    – lioness99a
    May 16 '17 at 14:50
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    $\begingroup$ The abscissas of the points of inflection are the values for which the sign of $f''(x)$ changes. So it is erroneous to consider the values where $f'(x)=0$. Only pay attention to those for which $f''(x)=0$. $\endgroup$
    – Jean Marie
    May 16 '17 at 15:11

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