3.3 Increasing and Decreasing Functions

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In this section, we use the derivative to determine intervals on which a given function is increasing or decreasing. We will also determine the local extremes of the function.

Increasing and Decreasing Functions

A function $f(x)$ is called increasing on an interval $I$ if given any two numbers, $x_1$ and $x_2$ in $I$ such that $x_1 < x_2$ , we have $f(x_1) < f(x_2)$ .

Similarly, $f(x)$ is called decreasing on an interval $I$ if given any two numbers, $x_1$ and $x_2$ in $I$ such that $x_1 < x_2$ , we have $f(x_1) > f(x_2)$ .

To determine where a function is increasing or decreasing, we use the derivative as in the following theorem.

Increasing/Decreasing If $f'(x) > 0$ on an interval $I$ , then $f(x)$ is increasing on $I$ , and if $f'(x) < 0$ on an interval $I$ , then $f(x)$ is decreasing on $I$ .

Determine intervals on which $f(x) = 2x^3 - 3x^2 - 36x + 2$ is increasing or decreasing.
According to the theorem, we must determine where $f'(x)$ is positive and where $f'(x)$ is negative. To do this, it is often easiest to first determine where $f'(x) = 0$ or $f'(x)$ is undefined. In this example, $f'(x) = 6x^2 - 6x -36$ which exists for all $x$ . We solve the equation $f'(x) = 6x^2 - 6x -36 = 0$ which yields $6(x+2)(x-3) = 0$ and hence, $x=-2$ or $x=3$ . Note that these are the critical numbers of $f(x)$ . These two $x$ -values break the real number line into three open intervals: $(-\infty , -2), (-2, 3)$ and $(3, \infty )$ . On each of these intervals, $f'$ will be either strictly positive or strictly negative. To determine which, we will use test points. For the interval, $(-\infty , -2)$ we will use $-3$ as the test point (any number in the interval is acceptable). Then we consider the derivative at this point: $f'(-3) = (-3)^2 - (-3) -6 = 6 >0$ . This means that $f'(x) > 0$ for every $x$ value in the interval $(-\infty , -2)$ and by the theorem, $f(x)$ is increasing on this interval.

Next, for the interval $(-2, 3)$ , we will use $0$ as the test point, and we have $f'(0) = (0)^2 - (0) -6 = -6 <0$ . This means that $f'(x) < 0$ for every $x$ value in the interval $(-2, 3)$ and by the theorem, $f(x)$ is decreasing on this interval.

Finally, for the interval $(3, \infty )$ , we will use $4$ as the test point and we have $f'(4) = (4)^2 - (4) -6 = 10>0$ . This means that $f'(x) > 0$ for every $x$ value in the interval $(3, \infty )$ and by the theorem, $f(x)$ is increasing on this interval.

Thus $f(x)$ is increasing on the intervals $(-\infty , -2)$ and $(3, \infty )$ , and it is decreasing on the interval $(-2, 3)$ .

The work that was done in the previous example can actually give us slightly more information about $f(x)$ . We can determine the local extremes of $f(x)$ .

Local Extremes We say that $f(x)$ has a local maximum at $x = a$ if there is an open interval $I$ containing $a$ such that $f(x) \leq f(a)$ for all values of $x$ in $I$ . Similarly, we say that $f(x)$ has a local minimum at $x = a$ if there is an open interval $I$ containing $a$ such that $f(x) \geq f(a)$ for all values of $x$ in $I$ .

Critical numbers can help us find the location and the nature of local extremes and the next theorem tells us how.

First Derivative Test If $f'(x)$ changes sign from positive to negative at the critical number $x=a$ , then $f(x)$ has a local maximum at $x = a$ and if $f'(x)$ changes sign from negative to positive at the critical number $x=a$ , then $f(x)$ has a local minimum at $x = a$ .

Find the local extremes of $f(x) =2x^3 - 3x^2 - 36x + 2$ .
In the previous example, we found that the critical numbers for $f(x)$ were $-2$ and $3$ . We also noted that $f'(x)$ changed sign from positive to negative at $x = -2$ and from negative to positive at $x=3$ . Hence, by the First Derivative Test, $f(x) = 2x^3 - 6x^2 - 36x + 2$ has a local maximum at $x = -2$ and a local minimum at $x=3$ . $\graph {2x^3 - 3x^2 - 36x + 2}$

Find the local extremes of the function $f(x) = x^2 - 6x + 2.$