Function increase or decrease The question is

$$\text{Let } f(r) = r^{1/3} + \frac 1r \text{ for } r>0$$
a) Determine where the function $f$ is increasing or decreasing.
b) Determine where the function $f$ is concave upward or downward.

I've worked out
$$f'(r)= \frac 13r^{-2/3} -\frac 1{r^2}$$
but don't know how to work out the values of $r$.
What do I do next?
 A: I'll help you with part a). We can see that $f'(r)$ is defined for all $r>0$, so we just need to find where $f'(r)=0$. Let's start by rewriting that as
$$\frac 13r^{-2/3} -r^{-2}=0$$
Move the second term to the other side to get
$$\frac 13r^{-2/3}=r^{-2}$$
Taking the reciprocal of each side, remembering that $\dfrac 1{a^{-b}}=a^b$,
$$3r^{2/3}=r^2$$
Take both sides to the third power to get
$$3^3r^2=r^6$$
Divide by $r^2$ to get
$$3^3=r^4$$
and finally
$$r=3^{3/4}=\sqrt[4]{27}$$
This is where the derivative is zero. Now look at the intervals $(0,3^{3/4})$ and $(3^{3/4},\infty)$ to find where the derivative is positive and where it is negative, i.e. where the function increases and where it decreases. Check all this with a graph:

Let me know if you need more help.
A: The function is increasing when the derivative 
is positive and decreasing when the derivative is negative.
The function is concave upwards when the second derivative is positive
and concave downward when the second derivative is negative. 
Set the first and second derivative to zero and work from there. 
$$f''(r) = \frac{2}{r^3} - \frac{2}{9r^{5/3}}$$
$$\frac{2}{r^3} - \frac{2}{9r^{5/3}} = 0 $$
