Is finding the second derivative of $\sqrt[3]{\vert x\vert}$ the best method to determine if it is convex?

Question

I have an exercise where I have to tell on which intervals a function is concave or convex. I usually do it using second derivative, but I would like to know if there is a simpler way of doing so, because this gets a little messy for me when considering $ \sqrt[3]{\vert x\vert } $.

Answer 1

HINT:

Draw the graph of $y=|x|^\frac{1}{3}$ for $x>0$ i.e. $y=x^\frac{1}{3}$ and reflect the graph about the y-axis to get the graph of $y=|x|^\frac{1}{3}$ for $x<0$.

Only you have to check the differentiablity of $f(x)$ at $x=0$.

Does it help?

Answer 2

Observe that $|x|=\sqrt{x^2}$

Now put $y^3=$$|x|=\sqrt{x^2}$

So $\displaystyle3y^2 \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{\{x^2\}^{\frac{1}{2}}}\cdot 2x =\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}$

So $\displaystyle\frac{dy}{dx}=\frac{x}{3y^2|x|}$