Identifying Inflection Points for This Strange Function: $y=\frac{3}{11}\left(x^2-1\right)^{\frac{2}{3}}$

Question

Identify the inflection points of $y=\frac{3}{11}\left(x^2-1\right)^{\frac{2}{3}}$.

The inflection points are $\left(\pm \sqrt{3},\frac{3}{11}\sqrt[3]{4}\right)$.

However, I'm not sure how to get this answer.

Can someone please explain it to me?

Answer 1

Differentiate twice you'll get $$\frac{4 (x^2 - 3)}{33 (x^2 - 1)^{\frac{4}{3}}}$$ Set it to zero you have values of $x$. For values of $y$, plug $x$'s into the function.

Edit:

By chain rule, you have $y'=\frac{2}{11}(x^2-1)^{\frac{-1}{3}} \times2x = \frac{4x}{11(x^2-1)^{\frac{1}{3}}}$.

Differentiate agian to obtain $y''$. Quotient rule this time: $$y''=[4\times 11 (x^2-1)^{\frac{1}{3}}-4x\times \frac{11}{3}(x^2-1)^{\frac{-2}{3}}\times2x]/[11^2(x^2-1)^{\frac{2}{3}}]$$ Multiply top and bottom by $(x^2-1)^{\frac{2}{3}}$.

$$y''=\frac{44(x^2-1)-\frac{88}{3}x^2}{11^2(x^2-1)^{\frac{4}{3}}}$$

Simplify a bit you get the answer.

Answer 2

Set $y''(x)=0$, and double-check that either $y'''(x)\not=0$ or $y^{(5)}(x)\not=0$, ... You want the lowest-order nonzero derivative beyond the second to be of odd order. Those are sufficient conditions to force an inflection point.

Answer 3

Are you sure of that coordinates? We have $$y'(x)=\frac{2}{11}(x^2-1)^{2/3-1}2x$$ and $$y'(x)=\frac{4}{11}\frac{x}{(x^2-1)^{1/3}}$$ and $$y''(x)={\frac {4\,{x}^{2}-12}{33\, \left( \left( x-1 \right) \left( x+1 \right) \right) ^{4/3}}} $$ Now solve the equation $$y''(x)=0$$ for $x$ and the coordinates in $$y'''(x)$$ Solving the equation $$y''(x)=0$$ we get $$x=\pm\sqrt{3}$$ and $$y=\frac{3}{11}2^{2/3}$$