Given $f'(x)=9x^2-4$ find the following:
$i)$ The open intervals where $f'(x)=9x^2-4$ is increasing and decreasing.
$ii)$ The open intervals where $f(x)$ is concave upwards and concave downwards
$iii)$ The x values where relative extrema of the function $f(x)$ occur
$iv)$ the x value where a point of inflection occurs
Solution:
$i)$ Remark: We are really good at finding where a function $g(x)$ is increasing and decreasing. You just take it's derivative, find the critical points, and test the sign of derivative in each region. But this question is asking where the DERIVATAIVE of some function is increasing and decreasing, but don't panic, because you do the exact same thing, except the derivative of the function will be the second derivative of some original function (that we are not given).
We want to find the open intervals where $f'(x)=9x^2-4$ is increasing and decreasing. Therefore, we have to take it's derivative to get $f''(x)$, (which will be a second derivative of some function $f(x)$ that we were not given), then find the critical points, draw number line and separate based on the critical points, then test the sign of $f''(x)$ in each region.
Cool. Lets take the derivative of $f'(x)$ and find the critical points
$f''(x)=18x=0$
$\rightarrow x=0$. Since $f''(x)$ exists everywhere, $x=0$ is our only critical point
Let's take a number less than $0$ and a number greater than zero to find the sign of $f''(x)$ in each of the regions around the critical point.
$f''(-1)=-18<0$
$f''(1)=18>0$
Therefore $f'(x)$ is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$
$ii)$ To find out where $f(x)$ is concave upward and downward, we have to find the critical points of the second derivative, make a number line, and then test the sign in each of the regions.
BUT WE ALREADY DID THIS IN PART 1! The key observation here is that a function is concave upwards where it's derivative is increasing, and a function is concave downwards where its derivative is decreasing.
$\rightarrow f(x)$ is concave downwards on $(-\infty,0)$ and concave upwards on $(0,\infty)$
$iii)$ The $x$ values where a relative extrema of $f(x)$ may occur.
These are the values of $x$ where $f'(x)=0$ or $f'(x)$ DNE
$f'(x)=9x^2-4=0$
$\rightarrow x = \pm \frac{2}{3}$
And $f'(x)$ exists everywhere so only $x$ values where relative extrema may occur are $x = \pm \frac{2}{3}$
$iv)$: The points of inflection are where the sign of $f''(x)$ changes. When we were finding where $f'(x)$ is increasing and decreasing (which is wherever $f(x)$ is concave upward and downward), we saw that the $f''(x)$ is negative to the left of $x=0$ and positive to the right of $x=0$. Therefore $x=0$ is a x value for a point of inflection.