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?
Best Answer
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 $(-\infty, \infty)$.