Question

Use the 1st derivative test to find the intervals where the function is increasing or decreasing. Please show your work so I can get a better understanding on how you achieved the solution. Thank you.

Answer 1

As the name of the test suggests, you have to compute the derivative:

`f(x) = \frac1{2x+3} \implies f'(x) = -\frac2{(2x+3)^2}`

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

`f(x) = (x+1)/(x+3) \implies f'(x) = ((x+3)-(x+1))/((x+3)^2) = \frac2{(x+3)^2}`

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.