Find the open interval on which $x^{1/3}-9$ is increasing/decreasing Consider the following function
$$f(x) = x^{\frac{1}{3}} - 9$$
How can I find the open interval on which the function is increasing or decreasing?
a. I know the critical points are $(0,-9)$
b. I know that the interval is not decreasing.
c. I know that by applying first derivative test to identify the relative extremum. (In this case no extrema).
All I cannot figure out is: in which interval the function is increasing? 
 A: To find where a function is increasing or decreasing, we look at the first derivative:
$$f(x) = x^{1/3} - 9\\
f'(x) = \frac{x^{-2/3}}{3}$$
We set the derivative equal to zero:
$$\frac{x^{-2/3}}{3} = 0$$
$$x^{-2/3} = 0$$
$$\frac{1}{x} = 0$$
Critical points come in two categories: stationary and singular.  Stationary points are where the first derivative is equal to zero; singular points are where it is undefined.
Note that $\frac{1}{x} \ne 0$ for any $x$; thus, there are no stationary points!
However, the function $f'(x)$ is undefined at $x=0$, so there is a singular point at $x=0$.
Now, we test $f'(x)$ to see if it is positive or negative on either side of the origin:
$$f(-1) = \frac{1}{3},\qquad f(1) = \frac{1}{3}$$
Thus, the first derivative is positive for all x values, except $x=0$ (where it is undefined).  This tells us that the function is increasing on the interval $(-\infty, 0)$ and on $(0, \infty)$.  But what about at $x=0$?  
As $f$ is continuous at $x=0$, $f$ is increasing on all of $\mathbb{R}$.  So, our final interval is $(-\infty, \infty)$.
A: Sara, your answer $(-\infty , 0)$ is not wrong, since it is true that the function is increasing in this interval. The same holds for $(0 , + \infty)$.
The answer must be $(- \infty , + \infty )$ only if the question was to find the largest interval where the function is increasing or decreasing :)
