increasing or decreasing nature of a...

Ogr. Gor. Volkan OGER FBA 1021 Calculus 1/ 46

Increasing or Decreasing Nature of a Function

Examining the graphical behavior of functions is a basic part ofmathematics and has applications to many areas of study. Whenwe sketch a curve, just plotting points may not give enoughinformation about its shape. For example, the points (-1,0), (0,-1),and (1,0) satisfy the equation given by y = (x + 1)3(x− 1).


On the basis of these points, we might hastily conclude that thegraph should appear as in Figure 1 (a), but in fact the true shapeis given in Figure 1(b).

Figure 1:


In this chapter we will explore the powerful role that differentiationplays in analyzing a function so that we can determine the trueshape and behavior of its graph.


We begin by analyzing the graph of the function y = f(x) inFigure 2. Notice that as x increases (goes from left to right) onthe interval I1, between a and b, the values of f(x) increase andthe curve is rising.

Figure 2:


Definition

A functionf is said to be increasing on an interval I when, for anytwo numbers x1, x2 in I if x1 < x2, then f(x1) < f(x2).

A function f is decreasing on an interval I when, for any twonumbers x1, x2 in I, if x1 < x2, then f(x1) > f(x2).


Turning again to Figure 2, we note that over the interval I1,tangent lines to the curve have positive slopes, so f ′(x) must bepositive for all x in I1. A positive derivative implies thai the curveis rising.


Over the interval I2, the tangent lines have negative slopes, sof ′(x) < 0 for all x in I2. The curve is falling where the derivativeis negative.


We thus have the following rule, which allows us to use thederivative to determine when a function is increasing or decreasing:

Rule ( Criteria for increasing or Decreasing Function)

Let f be differentiable on the interval (a, b). If

f ′(x) > 0 for all x ∈ (a, b)

then f is increasing on (a, b).

Iff ′(x) < 0 for all x ∈ (a, b)

then f is decreasing on (a, b).


To illustrate these ideas, we will use Rule to find the intervals on

which y = 18x− 2

3x3 is increasing and the intervals on which y is

decreasing. Letting y we must determine when f ′(x) is positiveand when f ′(x) is negative. We have

f ′(x) = 18− 2x2 = 2(9− x2) = 2(3 + x)(3− x)


We can find the sign of f(x) by testing the intervals determined bythe roots of 2(3 + x)(3− x) = 0, namely, −3 and 3. These shouldbe



These results are indicated in the sign chart, where the bottom lineis a schematic version of what the signs of f ′ say about f itself.

Notice that the horizontal line segments in the bottom rowindicate horizontal tangents for f at −3 and at 3.


Thus, f is decreasing on (−∞,−3) and (3,∞) and is increasingon (−3, 3).


This corresponds to the rising and falling nature of the graph of fshown in Figure 3.

Figure 3:Ogr. Gor. Volkan OGER FBA 1021 Calculus 15/ 46

Example

Example: Find where the function f(x) = 3x4 − 4x3 − 12x2 + 5 isincreasing and where it is decreasing..

Solution:

f ′(x) = 12x3 − 12x2 − 24x = 12x(x− 2)(x + 1)

To use the I/D Test we have to know where f ′(x) > 0 and wheref ′(x) < 0.

This depends on the signs of the three factors of f ′(x), namely,12x, x− 2 and x + 1.


Example...

We divide the real line into intervals whose endpoints are thecritical numbers −1, 0, 2 and arrange our work in a chart.

Decreasing Increasing Decreasing Increasing

A plus sign indicates that the given expression is positive, and aminus sign indicates that it is negative.


Example...

Therefore, the function f(x) = 3x4 − 4x3 − 12x2 + 5

is DECREASING on (−∞,−1),

is INCREASING on (−1, 0) ,

is DECREASING on (0, 2),

is INCREASING on (2,∞).


Look now at the graph of y = f(x) in Figure 4. Some observationscan be made. First, there is something special about the pointsP,Q, and R.

Figure 4:

Notice that P is higher than any other ”nearby” point on thecurve—and likewise for R.


The point Q is lower than any other ”nearby” point on the curve.


Since P,Q, and R may not necessarily be the highest or lowestpoints on the entire curve, we say that the graph of f has relativemaxima at a and at c, and has a relative minimum at b.

The function f has relative maximum values of f(a) at a and f(c)at c; and has a relative minimum value of f(b) at b. We also saythat (a, f(a)) and (c, f(c)) are relative maximum points and(b, f(b)) is a relative minimum point on the graph off.


Turning back to the graph, we see that there is an absolutemaximum (highest point on the entire curve) at a, but there is noabsolute minimum (lowest point on the entire curve) because thecurve is assumed to extend downward indefinitely.


Definition

A function f has a relative maximum at a if there is an openinterval containing a on which f(a) > f(x) for all x in theinterval. The relative maximum value is f(a).

A function f has a relative minimum at a if there is an openinterval containing a on which f(a) < f(x) for all x in theinterval. The relative minimum value is f(a).


Definition

A function f has an absolute maximum at a if f(a) > f(x) for allx in the domain of f . The absolute maximum value is f(a).

A function f has an absolute minimum at a, if f(a) < f(x) for allx in the domain of f . The absolute minimum value is f(a).


We refer to either a relative maximum or a relative minimum as arelative extremum (plural: relative extrema). Similarly, we speak ofabsolute extrema.


Referring to Figure 4, we notice that at a relative extremum thederivative may not be defined (as when x = c). But whenever it isdefined at a relative extremum, it is 0 (as when x = a and whenx = b) and hence the tangent line is horizontal. We can state thefollowing:

Rule (A Necessary Condition for Relative Extrema)

If f has a relative extremum at a, then f ′(a) = 0 or f ′(a) does notexist.


Rule does not say that if f(a) is 0 or f(a) does not exist, thenthere must be a relative extremum at a. In fact, there may not beone at all. For example, in Figure 5(a), f ′(a) is 0 because thetangent line is horizzontal at a, but there is no relative extremumthere.

Figure 5:


In Figure 5(b), f ′(a) does not exist because the tangent line isvertical at a, but again there is no relative extremum there.


Definition

For a in the domain of f , if either f ′(a) = 0 or f(a) does notexist, then a is called a critical value for f . If a is a critical value,then the point (a, f(a)) is called a critical point for f .

At a critical point, there may be a relative maximum, a relativeminimum, or neither.


Rule

Suppose f is continuous on an open interval I that contains thecritical value a and f is differentiable on I, except possibly at a.

1 If f ′(x) changes from positive to negative as x increasesthrough a, then f has a relative maximum at a.

2 If f ′(x) changes from negative to positive as x increasesthrough a, then f has a relative minimum at a.


Example

Test y = f(x) = x2ex for relative extrema.

By the product rule

f ′(x) = ex(2x) + x2ex = xex(x + 2)

Noting that ex is always positive, we obtain the critical values 0and −2.


From the sign chart of f(x) given in Figure 6, we conclude thatthere is a relative maximum when x = −2 and a relative minimumwhen x = 0.

Figure 6:


Example

Sketch the graph of y = f(x) = 2x2 − x4 with the aid ofintercepts, symmetry, and the first-derivative test.


Example

Test y = F (x) = x2/3 for relative extrema.

We have

f ′(x) =2

3x−1/3

=2

3 3√x

Since f ′(0) does not exist, x = 0 is critical point. And there is noreal number such that f ′(x) = 0.


So there is only one critical point. The sign chart is given in thefollowing figure:

Figure 7:

Since f(x) is defined at x = 0, f has a relative minimum at 0 off(0) = 0, and there are no other relative extrema.


Example

If c = 3q − 3q2 + q3 is a cost function, when is marginal costincreasing?


Example (Storage and Shipping Costs)

In his model for storage and shipping costs of materials formanufacturing process, Lancaster derives the cost function

C(k) = 100

(100 + 9k +

144

k

)1 ≤ k ≤ 100

Where C(k) is the total cost (indollars) of storage andtransportation for 100 days of operation if a load of k tons ofmaterial is moved every k days.

1 Find C(1)

2 For what values of k does C(k) have a minimum?

3 What is the minimum value?


Absolute Extrema on a Closed Interval

Theorem (Extreme-Value Theorem)

If a function is continuous on a closed interval, then the functionhas both a maximum value and a minimum value on that interval.


For example, each function in Figure 8 is continuous on the closedinterval [1, 3]. Geometrically, the extreme-value theorem assures usthat over this interval each graph has a highest point and a lowestpoint.

Figure 8:


We will focus our attention on absolute extrema and make use ofthe extreme-value theorem where possible.

If the domain of a function is a closed interval, to determineabsolute extrema we must examine the function not only at criticalvalues, but also at the endpoints.


For example. Figure 9 shows the graph of the continuous functiony = f(x) over [a, b]. The extreme-value theorem guaranteesabsolute extrema over the interval. Clearly, the important pointson the graph occur at x = a, b, c, and d, which correspond toendpoints or critical values.

Figure 9:Ogr. Gor. Volkan OGER FBA 1021 Calculus 41/ 46


Example

Find absolute extrema for f(x) = x2 − 4x + 5 over the closedinterval [1, 4]

To find the critical values of f , we must find f ′

f ′(x) = 2x− 4 = 2(x− 2)

this gives the critical value x = 2.

Evaluating f(x) at the end points 1 and 4 and at the critical value2, we have

f(1) = 2 f(4) = 5 f(2) = 1

From the values, we conclude that the maximum is f(4) = 5 andthe minimum is f(2) = 1


Example

Find absolute extrema for f(x) = −3x5 + 5x3 over the closedinterval [−2, 0]


Example (Maximizing Revenue)

The demand equation for a manufacturer’s product

p =80− q

40 ≤ q ≤ 80

where q is the number of units and p is the price per unit. Atwhich value of q will there be maximum revenue? What is themaximum revenue?


Example

Suppose that the demand equation for a monopolist’s product isp = 400− 2q and the average-cost function isc = 0.2q + 4 + (400/q). where q is number of units, and both pand c are expressed in dollars per unit.

1 Determine the level of output at which profit is maximized.

2 Determine the price at which maximum profit occurs.

3 Determine the maximum profit.

4 If, as a regulatory device, the government imposes a tax of$22 per unit on the monopolist, what is the new price forprofit maximization?


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