# We have f:RR->RR,f(x)=|x|root(3)(1-x^2).How to find maximum domain of differentiability?

May 24, 2017

See below.

#### Explanation:

$f \left(x\right) = | x | \sqrt[3]{1 - {x}^{2}} = \left\{\begin{matrix}x \sqrt[3]{1 - {x}^{2}} & x \ge 0 \\ - x \sqrt[3]{1 - {x}^{2}} & x < 0\end{matrix}\right.$

$\frac{d}{\mathrm{dx}} \left(x \sqrt[3]{1 - {x}^{2}}\right) = \sqrt[3]{1 - {x}^{2}} + x \left(\frac{1}{3} {\left(1 - {x}^{2}\right)}^{- \frac{2}{3}} \left(- 2 x\right)\right)$

$= \sqrt[3]{1 - {x}^{2}} - \frac{2 {x}^{2}}{3 {\left(1 - {x}^{2}\right)}^{\frac{2}{3}}}$

$= \frac{3 \left(1 - {x}^{2}\right) - 2 {x}^{2}}{3 {\left(1 - {x}^{2}\right)}^{\frac{2}{3}}}$

$= \frac{3 - 5 {x}^{2}}{3 {\left(1 - {x}^{2}\right)}^{\frac{2}{3}}}$. $\text{ }$ For $x \ne \pm 1$

So,

$f ' \left(x\right) = \left\{\begin{matrix}\frac{3 - 5 {x}^{2}}{3 {\left(1 - {x}^{2}\right)}^{\frac{2}{3}}} & x > 0 & x \ne 1 \\ - \frac{3 - 5 {x}^{2}}{3 {\left(1 - {x}^{2}\right)}^{\frac{2}{3}}} & x < 0 & x \ne - 1\end{matrix}\right.$
And $f$ is not differentiable at $\pm 1$

Checking the "joint" of the "hinge" of the two parts, we see that the right derivative at $0$ is $1$ and the left derivative at $0$ is $- 1$, so there is no derivative at $0$.

Finally, here is the graph of the function: graph{y=(1-x^2)^(1/3)absx [-2.433, 2.436, -1.215, 1.217]}