As the name of the test suggests, you have to compute the derivative:
Find where, if at all, the derivative vanishes or is undefined - these are the critical points of <em>f(x)</em>.
In this case, the derivative is never 0 since the numerator is constant and the denominator is non-negative. You also see that <em>f '(x)</em> is negative over its entire domain.
The denominator goes to 0 when 2<em>x</em> + 3 = 0, or <em>x</em> = -3/2.
Now split up the domain of <em>f(x)</em> into intervals with endpoints at the critical points. Here, we consider the two intervals, (-∞, -3/2) and (-3/2, ∞).
Take any point from either interval and check the sign of <em>f '(x)</em> at that point. Any points will do, but you should strive to pick one that makes calculations simple.
• From (-∞, -3/2), take <em>x</em> = -2; then <em>f '</em> (-2) = -2 < 0. Since <em>f '(x)</em> is negative over this interval, <em>f(x)</em> is decreasing over it.
• From (-3/2, ∞), take <em>x</em> = 0; then <em>f ' </em>(0) = -2/9 < 0. Again, this means <em>f(x)</em> is decreasing over this interval.
So, the first derivative test tells us that <em>f(x)</em> = 1/(2<em>x</em> + 3) is decreasing over the intervals (-∞, -3/2) and (-3/2, ∞); in other words, over its entire domain.
For the second function, we have
Again, there's only one critical point, this time at <em>x</em> = -3 where the derivative is undefined.
• From the interval (-∞, -3), take <em>x</em> = -4; then <em>f '</em> (-4) = 2 > 0, so <em>f(x)</em> is increasing.
• From the interval (-3, ∞), take <em>x</em> = 0; then <em>f '</em> (0) = 2/9 > 0, so <em>f(x)</em> is increasing.