4.1 Increasing/decreasing criterion

Here are a few definitions to begin with.

Informally, a function is increasing on an interval if its graph slopes up on that interval, and is decreasing on an interval if its graph slopes down on the interval. For example, the function is increasing on the interval , decreasing on the interval , and increasing on the interval , as illustrated in Figure 37.

Figure 37 The graph of the function

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The graph of the function f of x = 2 x cubed minus 3 x squared minus 36 x. The x-axis is marked from before minus 6 to beyond 6 in intervals of 2. There is a hollow dot on the axis at x = minus 2 and another at x = 3. The y-axis is marked from before minus 150 to beyond 150 in intervals of 50. The curve starts very low down, where both coordinates are negative. It rises to cut the negative x-axis between x = minus 4 and x = minus 3. It continues to rise until it turns at (minus 2, 44). It then falls to pass through the origin and continues to fall where the x coordinate is positive but the y coordinate is negative. It turns again at (3, minus 81) and then rises, cutting the x-axis again just after x = 5. The graph has a coloured background. There is central green rectangle which extends from the top of the graph to the bottom. It lies between x = minus 2 and x = 3, that is where the curve is decreasing. The remaining background, where the curve is increasing, is coloured mauve.

Figure 37 The graph of the function

Of course, you can’t tell from the graph alone that this function stops increasing and starts decreasing when is exactly , or that it stops decreasing and starts increasing when is exactly . You’ll see shortly how to confirm that the function is increasing and decreasing on the intervals mentioned above, exactly.

You could also say that the function in Figure 37 is increasing on the interval , decreasing on the interval , and increasing on the interval . However, when we discuss intervals on which a function is increasing or decreasing, it’s often helpful to consider intervals that don’t overlap, so we usually consider open intervals.

Since positive gradients correspond to a graph sloping up, while negative gradients correspond to a graph sloping down, the derivative of a function tells you the intervals on which the function is increasing or decreasing, as set out below.

For example, Figure 38 shows parts of the tangents to some points on the graph of the function . It illustrates that positive gradients correspond to intervals on which the graph is increasing, and negative gradients correspond to intervals on which the graph is decreasing.

Figure 38 Some tangents to the graph of

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Some tangents to the graph of f of x = 2 x cubed minus 3 x squared minus 36 x. This shows the same graph as in figure 37. The x-axis is marked from before minus 6 to beyond 6 in intervals of 2. The y-axis is marked from before minus 150 to beyond 150 in intervals of 50. The curve starts very low down, where both coordinates are negative. It rises to cut the negative x-axis between x = minus 4 and x = minus 3. It continues to rise until it turns at (minus 2, 44). It then falls to pass through the origin and continues to fall where the x coordinate is positive but the y coordinate is negative. It turns again at (3, minus 81) and then rises, cutting the x-axis again at about x = 5. The graph has a coloured background. There is central green rectangle which extends from the top of the graph to the bottom. It lies between x = minus 2 and x = 3, that is where the curve is decreasing. The remaining background, where the curve is increasing, is coloured mauve. Four points are marked on the graph and tangents are drawn at each of them. Two are where the graph is increasing. One has an x coordinate of about minus 5 and the other has an x coordinate of just over plus 5. The tangent at each of these points has a positive slope. Two points and their associated tangents are shown where the curve is decreasing. The points are at about x = minus 1 and x = 2.5 . Each of these tangents has a negative slope.

Figure 38 Some tangents to the graph of

It’s sometimes useful to determine whether a particular function that you’re working with is increasing on a particular interval, or decreasing on a particular interval. (For example, this can help you sketch its graph.) You can often use the increasing/decreasing criterion to do this, as illustrated in the next example. This example confirms that the function in Figure 37 is increasing and decreasing on the intervals mentioned earlier.

Activity 13 Using the increasing/decreasing criterion

Consider the function .

• a.Find the derivative , and factorise it.

• b.Show that is increasing on each of the intervals and .

• c.Show that is decreasing on the interval .

Answer

• a., so

• b.When is less than , the values of and are both negative, and hence the value of is positive.

  Similarly, when is greater than 5, the values of and are both positive, and hence the value of is also positive.

  Therefore, by the increasing/decreasing criterion, the function is increasing on each of the intervals and .

• c.When is in the interval , the value of is positive and the value of is negative, and hence the value of is negative.

  Therefore, by the increasing/decreasing criterion, the function is decreasing on the interval .

(The graph of is as follows.)

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The x-axis is marked from zero to beyond 7 in intervals of 1. The y-axis is marked from below minus 6 to above 6 in intervals of 2. The graph of y = two thirds of x cubed minus 8 x squared + 30 x minus 36 is shown. It starts low down, with x at about 1.5 and rises to touch the x axis at (3, 0). It immediately falls and then turns again at x = 5. It then rises, cutting the x-axis at x = 6. It continues to rise.

An alternative way to set out working of the type in Example 6 and in the solution to Activity 13 is to use a table of signs. You’ll see examples of this in the next section.

Activity 14 Using the increasing/decreasing criterion again

Consider the function .

• a.Find the derivative , and complete the square on this expression.

• b.Hence show that is increasing on the interval (that is, on the whole of its domain).

Answer

• a., so

• b.For every value of , the expression is non-negative, and hence the expression is positive.

  That is, the derivative is positive on the interval . Therefore, by the increasing/decreasing criterion, the function is increasing on the interval .

(The graph of is shown below.)

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The x-axis is marked from before minus 5 to beyond 5 in intervals of 1. The y-axis is marked from before minus 80 to beyond 80 in intervals of 20. The graph of y = x cubed minus 3 x squared + 4 x + 3 is shown. It starts low down where both coordinates are negative. It rises, begins to flatten out as it cuts the x-axis at about minus 0.5 and continues fairly flat, cutting the y-axis at (0, 3). It begins to rise more steeply when x is greater than 2.

If you look back to the statement of the increasing/decreasing criterion, you’ll see that the first part begins

A slightly more concise way to express the same thing is to say

Similarly, the beginning of the second part of the increasing/decreasing criterion,

can be expressed as

These more concise forms are used in the next section.