Process for finding intervals of increase/decrease

Tips

• The intervals we will consider are the intervals between the critical numbers on the $x$-axis, and between the smallest critical number and negative infinity (if $f$ is defined on that interval) and similarly between the largest $c$ and positive infinity.
• To check on the sign of $f'$ on an interval, one can pick a number $b$ (a favorite, easy number), and find the sign of each factor of $f'$ at that number. Then, using what we know about the products of positive and negative numbers, we can find the sign of $f'(b)$.
• If $f'(b)$ is positive, the function is increasing on that entire interval.  This is because $f'(x)$ is positive at $x=b$, and can't change sign anywhere else in the interval (similarly if $f'(b)$ is negative).  Remember: $f'(x)$ can only change sign at critical points! 

        Graph of $f$:                Graph of $f'$:       

DO:  Try to follow the process (above) to work this problem before looking at the solution below.


Solution:

1. $f'(x)=3x^2-6x=3x(x-2)$
   
   
2. Since $f'$ is always defined, the critical numbers occur only when $f'=0$, i.e., at $c=0$ and $c=2$.


3. Our intervals are $(-\infty,0)$, $(0,2)$, and $(2,\infty)$.


4. On the interval $(-\infty,0)$, pick $b=-1$.  (You could just as well pick $b=-10$ or $b=-0.37453$, or whatever, but $-1$ is simplest.)  Both factors are negative, so $f'(-1)$ is positive.  (You can also get this by just evaluating $f'(-1)=9$).  On the interval $(0,2)$, pick $b=1$.  One factor ($x$) is positive, while the other ($x-2$) is negative, so the product is negative.  Or just evaluate $f'(1)=-3$.  On the interval $(2,\infty)$, pick $b=3$. Both factors are positive, so $f'(3)$ is positive.  Note that all we need here is the sign of $f'$, not its value.


5. So our function is increasing on $(-\infty,0)$, decreasing on $(0,2)$, and then increasing again on $(2,\infty)$.