Section 3.1 #11:
Find the critical numbers of $g(x)=(x-3)^{2/3}$ and find the open intervals where the function is increasing or decreasing.
$Solution$:
The critical points are the values of $x$ at which either $g'(x)=0$ or $g'(x)$ DNE. So, first we have to find $g'(x)$, and it looks like we are going to have to use the chain rule to do it.
$g'(x) = \frac{d}{dx} (x-3)^{2/3} = \frac{2}{3}(x-3)^{\frac{-1}{3}} = \frac{2}{3(x-3)^{\frac{1}{3}}}$
Okay, first lets find where
$g'(x)= \frac{2}{3(x-3)^{\frac{1}{3}}}=0$
Multiplying both sides of this equation by the denominator:
$(3(x-3)^{\frac{1}{3}})\frac{2}{3(x-3)^{\frac{1}{3}}}=0(3(x-3)^{\frac{1}{3}})$
$\rightarrow$
$2=0$.
Well shoot, that's not right. It looks like the derivative of $g(x)$ never equals zero.
What about the values of $x$ that are critical points because $g'(x)$ DNE? Well, that would be values of $x$ that make the denominator zero, and so we must have $x=3$.
So $x=3$ is our only critical point.
Now we split the number line up into two different sections, all the numbers to the left of $3$ and then all the numbers to the right of $3$, we pick a number from each region and plug it into the derivative to see if the original function of $g(x)$ is increasing or decreasing there.
$g'(0) = \frac{2}{3(0-3)^{\frac{1}{3}}} = \frac{2}{3(-3)^{\frac{1}{3}}}<0$
Note we don't have to know exactly what this number equals, only whether it's positive or negative.
So, since $g'(0)$ was negative, we have $g(x)$ is decreasing on $(-\infty,3)$
Now lets plug in a number to the right of $3$... how about 5.
$g'(5) = \frac{2}{3(5-3)^{\frac{1}{3}}} = \frac{2}{3(3)^{\frac{1}{3}}}>0$
Therefore $g(x)$ is increasing on $(3,\infty)$