First Derivative Sign ChartsA function f(x) is said to be increasing if

f(x1) ≤ f(x2) wheneverx1 ≤ x2,

and it is said to be decreasing if

f(x1) ≥ f(x2) wheneverx1 ≤ x2.

The derivative of a function gives us an efficient way to identify wherea function is increasing and decreasing:

If f ′(x) > 0 on an interval, then f is increasing there.If f ′(x) < 0 on an interval, then f is decreasing there.

Example 1: Let f(x) = x3 − 48x. Then f ′(x) = 3x2 − 48. Thederivative is zero when

3x2− 48 = 0 =⇒ x2

− 16 = 0 =⇒ x = ±4.

For x in the interval (−4, 4), we see that f ′(x) < 0, so f is decreasing on(−4, 4). Since f ′(x) > 0 on (−∞,−4) and on (4,∞), f is increasing onthose intervals. �

The analysis in the last example is confirmed by a graph of the functionf(x) = 3x2 − 48x:

When we are analyzing the sign of f ′ in order to detect where f isincreasing and decreasing, it is useful to have a good way of organizingthis information. Once such way is to graph a number line, representingthe values of x in the domain of f , and then above to number line toindicate the corresponding sign of f ′ – positive, negative, or zero. For thefunction in Example 1 above, we would obtain the following:

-4 4

+++---+++ 00

x

f’(x)

This is called a first-derivative sign chart. The chart indicatesthat f ′ is positive on (−∞,−4) and on (4,∞), and that it is negativeon (−4, 4). In general, once we have found the critical points (the

1

2

values of x where f ′′(x) = 0 or where the derivative is undefined), we candetermine the appropriate sign on each interval by testing points. For thelast example, we could have seen that f ′(0) = −48, and therefore f ′ isneagtive on (−4, 4).

Example 2: Use a first-derivative sign chart to find theinterval(s) where f(x) = 3x4 − 8x3 − 90x2 is increasing.

First we find the derivative:

f ′(x) = 12x3− 24x2

− 180x

Then we find the critical points (where f ′(x) = 0):

12x3− 24x2

− 180x = 0

=⇒ 12x(x2− 2x − 15) = 0

=⇒ 12x(x − 5)(x + 3) = 0

=⇒ x = 0, x = 5 or x = −3

Now we can draw our first-derivative sign chart:

-3 5

+++--- +++ 00

x

f’(x)

0

0---

(An easy way to determine which sign belongs to each interval is bytesting the values of f ′(−4), f ′(−1), f ′(1) and f ′(5). For example,f ′(1) = −192, and that is why there are negative signs above the interval(0, 5).)

Based on this chart, we conclude that f is increasing on the intervals(−3, 0) and (5,∞). �