Monotonicity and the Sign of the Derivative

Looking at the graph of a function, we usually see parts where the graph is "going up" and other parts where it is "going down". These behaviors can be made precise in the following way: We say that a function f(x) is increasing on an interval Iiff for any $x_1, x_2 \in I$

$$x_1 \leq x_2 \;\;\mbox{implies}\;\; f(x_1) \leq f(x_2)\;.$$

Similarly, we will say that f(x) is decreasing on I iff for any $x_1, x_2 \in I$

$$x_1 \leq x_2 \;\;\mbox{implies}\;\; f(x_2) \leq f(x_1)\;.$$

The increasing behavior corresponds to the up-motion seen on the graph, while the decreasing behavior corresponds to the down-motion.

These two properties of a function are closely related to the behavior of the derivative of the function (when it is differentiable). Indeed, it is quite easy to see that f(x) is increasing on I iff for any $x_1, x_2 \in I$

$$\frac{f(x_1) -f(x_2)}{x_1-x_2} \geq 0 \;,$$

and f(x) is decreasing on I iff for any $x_1, x_2 \in I$

$$\frac{f(x_1) -f(x_2)}{x_1-x_2} \leq 0 \;.$$

Using the Mean Value Theorem and the above ideas, we get the following powerful result:

Let f(x) be a continuous function on an arbitrary interval I and differentiable on its interior. Then

(1)

$f'(x) \geq 0$ for all x in the interior of I exactly if f(x) is increasing on all of I;

(2)

$f'(x) \leq 0$ for all x in the interior of I exactly if f(x) is decreasing on all of I;

(3)

f'(x) = 0 for all x in the interior of I exactly if f(x) is constant on all of I.

Remark. The practical difficulty you may encounter in using this result stems from the fact that you have to solve inequalities to determine the sign of the derivative. Consult our section on Solving Inequalities if you run into problems.

Here is a piece of advice: If the derivative is a product of functions, use the product rule for the derivative rather than multiplying out first!

Example. Consider the function

$$f(x) = 2x^3 - 3x^2 - 12 x + 1\;.$$

f(x) is a polynomial function. Therefore it is continuous and differentiable everywhere. Taking the derivative we get

$$f'(x) = 6x^2 - 6x - 12 = 6 (x-2)(x+1)\;.$$

So we have

$$\left\{\begin{array}{llll} f'(x) > 0 &\mbox{if $x < -1$}\\ f... ... f'(x) = 0 &\mbox{iff $x = -1$ or $x=2$}.\\ \end{array}\right.$$

So f(x) is increasing on the intervals $(-\infty,-1]$ and $[2,\infty)$, and f(x) is decreasing on the interval [-1,2].

From the graph, we see that the points x=-1 and x=2 are special. Indeed, at x=-1 the function behaves like a point at the top of a hill while at x=2 the graph looks like a valley. We will discuss this more on the next pages.

Exercise 1. Find the intervals on which $f(x) = x+ \sin(x)$is increasing or decreasing.

Answer.

Exercise 2. Find the intervals on which

$$f(x) = \left(\frac{1- \sqrt{x}}{1+x}\right)^7$$

is increasing or decreasing.

Answer.

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