f vs. f'

If possible, factor $f'$. If $f'$ is a quotient, factor the numerator and denominator (separately). This will help you find the sign of $f'$.

For each interval (in between the critical number tick marks) in which the function $f$ is defined, pick a number $b$, and use it to find the sign of the derivative $f'(b)$.

If $f'(b) > 0$, draw a straight line slanting upward over that interval on your number line. Similarly, if $f'(b) < 0$, draw a straight line slanting downward.

That's it! You can now see the intervals where $f$ is increasing and decreasing.

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!

DO: For any $b$ in an interval as above, $f'(b)$ will never be 0, and $f'(b)$ will always be defined. Why?

Example: Find where $f(x) = x^3 -3x^2$ is increasing/decreasing.

Graph of $f$:

Graph of $f'$:

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

Solution:

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

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.

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