As the name of the test suggests, you have to compute the derivative:
[tex]f(x) = \dfrac1{2x+3} \implies f'(x) = -\dfrac2{(2x+3)^2}[/tex]
Find where, if at all, the derivative vanishes or is undefined - these are the critical points of f(x).
In this case, the derivative is never 0 since the numerator is constant and the denominator is non-negative. You also see that f '(x) is negative over its entire domain.
The denominator goes to 0 when 2x + 3 = 0, or x = -3/2.
Now split up the domain of f(x) 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 f '(x) at that point. Any points will do, but you should strive to pick one that makes calculations simple.
• From (-∞, -3/2), take x = -2; then f ' (-2) = -2 < 0. Since f '(x) is negative over this interval, f(x) is decreasing over it.
• From (-3/2, ∞), take x = 0; then f ' (0) = -2/9 < 0. Again, this means f(x) is decreasing over this interval.
So, the first derivative test tells us that f(x) = 1/(2x + 3) is decreasing over the intervals (-∞, -3/2) and (-3/2, ∞); in other words, over its entire domain.
For the second function, we have
[tex]f(x) = \dfrac{x+1}{x+3} \implies f'(x) = \dfrac{(x+3)-(x+1)}{(x+3)^2} = \dfrac2{(x+3)^2}[/tex]
Again, there's only one critical point, this time at x = -3 where the derivative is undefined.
• From the interval (-∞, -3), take x = -4; then f ' (-4) = 2 > 0, so f(x) is increasing.
• From the interval (-3, ∞), take x = 0; then f ' (0) = 2/9 > 0, so f(x) is increasing.
