4.3 How derivatives affect the shape of a graph. The first derivative test and the second derivative test.

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1 Chapter 4: Applications of Differentiation In this chapter we will cover: 41 Maximum and minimum values The critical points method for finding extrema 43 How derivatives affect the shape of a graph The first derivative test and the second derivative test 45 Summary of curve sketching 47 Optimization problems 41 Maximum and minimum values The critical points method for finding extrema Motivation: Finding extrema (maximum and minimum) values of a function is an extremely important application in mathematics (in particular, in our class it is an application of the derivative) For example, problems of maximizing profit, minimizing cost, maximizing (or minimizing) areas or volumes and many other similar problems (see for example, the problems in section 47) are such typical applications of the methods which we will learn in this chapter An entire branch of mathematics called optimization is dedicated to finding extrema of functions Goals: define points of extrema (local and global) ; develop the critical point method for finding global extrema of a function; I Definition of points of extrema The extreme value theorem: Questions: What are points of extrema for a function? What are maximum and minimum points for a function? Do all functions have extrema (over an interval )? Definition 1: A Global (absolute) extrema: Consider a function f : D R and c D Then f ( is a: global (or absolute) maximum value of ( global (or absolute) minimum value of ( Example 1: f on D if (1) f f ( for xd ( ; f on D if () f f ( for xd ( ; For example, in Figure 1, we see that the global maximum of f ( is 5 achieved for c 3and its global minimum is achieved for c 6 B Local extrema: f ( is a: 1

2 local maximum value of f ( on D if 0such that ; local minimum value of f ( on D if 0such that ; Example : a (3) f f ( for xc a, c a ( (that is, near a (4) f f ( for xc a, c a ( (that is, near For example, in Figure, we see that f ( has a global maximum at d : f (d), a global minimum at a: f (a), and local minima at c and e, and local maxima at b and d Note that, in general, a global maximum is the greatest local maximum (except when it is an endpoint) and a global minimum is the smallest local minimum (again, except when it is an endpoint) See also Figure 3 where a few local minima and maxima are displayed Note again that endpoints (of the interval where the function is considered) can be global extrema, but not local extrema, since they would not satisfy a condition of the type (3) (or (4)) Therefore, local extrema are always inside the interval Figure 4: Consider also Figure 4, where the function f 4 3 ( 3x 16x 18x is graphed on the interval 1,4 For this example, identify: local minima: local maxima: global minimum: global maximum: Now that we can identify from a graph points of extrema, let us try to answer the following essential question: Do all functions defined over a certain interval have a local (or global) maximum and a local (or global) minimum over that interval?

3 Example : f :,,, given by : f ( x what are its local minima and maxima? a) Let 3 b) Let 3 f : 1,1 1,1, given by : f ( x what are its local minima and maxima? Let 3 f : 1,1 1,1, given by : f ( x what are its local minima and maxima? f :, 0,, given by : f ( x what are its local minima and maxima? d) Let We see that some of these functions (such as the function given in have local extrema, while (most of) the others do not The following important theorem establishes which functions are guaranteed to have an extrema over their corresponding interval Theorem 1 (Extreme value theorem): Consider a continuous function f a, b R minimum value f (d) at some points c d a, b : Then f ( achieves a global maximum value f ( and a global, This theorem is given without proof, since the proof is quite advanced (based on Cantor s completeness axiom) Draw a few typical figures of a continuous function on a closed interval to convince yourself of its validity Then look again at the functions in Example in the light of this theorem Remark 1: 1 Note that we need two conditions to be verified in order for global extrema to exist: the function f ( needs to be continuous and the interval where the function is considered needs to be closed Otherwise, we may well have that the function does not have a (local or global) extremum on the interval ; The functions for which we want to find extrema in this chapter almost always satisfy these two conditions (that is they are continuous functions considered on a closed interval) and therefore global extrema are guaranteed to exist for these functions Figure 5: A function which satisfies the EVT Figure 6: A function which does not satisfy the conditions of the conditions of EVT: it has a global minimum but no global maximum, this is in agreement with the theorem 3

4 II Fermat s theorem Finding points of global extrema for a function: Therefore, so far we have learned how to identify the local and global extrema, and conditions under which such extrema are guaranteed to exist Next, we consider the problem of finding such extremum values for a given function (expression) for which the graph is not given Note (see Figure 10 in the textbook) that for most functions local extrema occur at stationary points of the function (that is points where f '( 0 ) although this is not always the case However, these points are good candidates for extrema, as the following theorem states: Theorem (Fermat): If f : a, b R has a local maximum or a local minimum at c a, b then f '( 0 So: If f : a, b R and c a, b and f ( is a local extrema and f '( exists, then f '( 0 (If interested, see the proof of this theorem in the textbook) and if f '( exists, Note that f '( 0 does not in general imply that f ( is a local extrema as Example (a to showed However, stationary points of f ( are valuable candidates for extrema of f ( Example 3: Consider : 1,1 0,1 f given by: f ( x In this case, the local extrema are again not points where f '( 0 (0 is the local minimum and f '(0) does not exist) Remark : However, for a function f : a, b R which is continuous on b c a, b and f '( exists f '( 0 (from Fermat s theorem) ; OR c a, b and f '( does not exist (as in example 3 above) ; OR c is either a or b (the endpoints of the interval ) a,, if c [ a, b] is an extrema, then either: These 3 possibilities are the only 3 possibilities for an extrema of a function f a, b R : This is the basis of the following definition and of the Critical point method which follows Definition (critical points) : A critical point (read a valuable candidate of an extrema) of a function f : a, b R is a value c a, b such that either f '( 0 or f '( does not exist Note, that after we find the critical values of a given function, the only other candidates of an extrema for the function are the endpoints of the interval 4

5 3/5 Example 4: Consider f ( x 4 x Determine the critical points of f ( The reasoning outlines in Remark above gives us the procedure of finding the global extrema of any continuous function defined on a closed interval: f : a, b R This procedure is oulined in the Critical Point Method below: Critical Point Method for finding the global extrema of a continuous function f a, b R : : Step 1: Find all critical values of f ( (that is values c where either f '( 0 or f '( does not exist) Find f ( for all of these critical values Step : Find f ( a) and f ( b) Step 3: The largest of the values calculated in Steps 1 and is the global maximum of ( the smallest of these values is the global minimum of ( f on b a, f on b a, ; Example 5: 3 1 a) Find the absolute extrema of f ( x 3x 1 for x 4 Do problems 47 and 48 from Exercise set 41 in the textbook 5

6 43 How derivatives affect the shape of a graph The first derivative test and the second derivative test Goals: prove the First Derivative Test and learn how to use it to find all local extrema of a function f ( on a given interval; prove the Second Derivative Test and learn how to use it to find all local extrema of a function f ( on a given interval; I The first derivative test : In order to develop this test, we pose the question: What does f '( say about f (? To answer this question, remember that in section 30 we introduced f '( as the slope of the tangent line to G f at x The basis of the first derivative test is given by the following: Theorem 1: Consider f ( differentiable on an interval I a) If f '( 0 on I then ( b) If f '( 0 on I then ( f is strictly increasing on I (that is : x x f x f x any x x I 1 1 1, for ) f is strictly decreasing on I (that is : x x f x f x any x x I 1 1 1, for ) Proof: Follows easily using the mean value theorem from f ( Use also your visual intuition to see why this theorem must be true This theorem allows us to construct a table of monotonicity for a function f (, as long as we can determine the sign of its derivative This table of monotonicity will allow us to find easily all local and global extrema of a function f ( on an interval I Example 1: 4 3 Consider f ( 3x 4x 1x 5 on I [3,3] Find where f ( is increasing and where it is decreasing on I, then construct a monotonicity table and use this table to determine all points of extrema of f ( on I The method shown in Example 1 suggests the following theorem: Theorem (The First Derivative Test): Consider f : D R, f ( continuous on D, and let cd a critical number of f ( Then : a) If f '( changes sign at c from positive to negative (as we move from left to right) then f ( has a local maximum at c b) If f '( changes sign at c from negative to positive (as we move from left to right) then f ( has a local minimum at c 6

7 Proof: Is immediate, if we construct a monotonicity table for f ( around c Example : Do problems 9 and 10 in the Exercise set 43 II What f ''( say about f (? Consider the following two representative graphs: Figure 1: a concave up function (Figure 1a) and a concave down function (Figure 1b) Definition 1: a) If f '( is increasing on some interval I (so if f ''( 0 on I), then f ( is called concave up on I; b) If f '( is decreasing on some interval I (so if f ''( 0on I), then f ( is called concave down on I A typical shape of a concave up function is (as shown in Figure 1a) For this type of function, f '( is increasing (so f ''( 0 ) A typical shape of a concave down function is (as shown in Figure 1b) For this type of function, f '( is decreasing (so f ''( 0) Look also at Figure 7 in the textbook and determine the regions where f ( is concave up and where it is concave down Convince yourself that on these regions f '( has the expected monotonicity behavior (so f ''( has the expected sign) Therefore, the sign of f ''( provides the concavity of the function f ( Definition : A point P( x, f ( ) on G is called an inflection point of a continuous function f ( if f ( changes concavity at P f (or, in other words, if f ''( changes sign at P) Therefore, a point where f '( changes sign is a point of local extremum for f (, and a point where f ''( changes sign is a point of inflection for f ( 7

8 Example 3: Draw a monotonicity table and a concavity table (that is a table which shows the sign of f ''( on D f ) for the function f ( x 4 3 x 4 Use this information to sketch the graph of f ( There is also a second derivative test for finding the extrema of functions However, it works only for stationary points and will not be covered in this class Comparison of the methods for finding extrema: So far we have developed the main methods for finding the points of extrema of a function f ( : the critical point method (in section 41) and the first derivative test (Theorem in section 43) A comparison of the advantages and disadvantages of these tests reveal that the first derivative test is usually the best (preferred) test to use when determining the points of extrema of a function since: For the Critical Point Method, only the global extrema are found No information is known about the other critical points (that is, if they are local extrema or no extrema at all) Also, this method is quite tedious and long The First Derivative Test determine all local extrema and the entire monotonicity behavior of f ( on I It is the best since it provides a complete information about the monotonicity of the function f ( if we can determine the sign of f '( ; Therefore, when possible, it is preferred to use the First Derivative Test when determining the extrema of a function Example 5: Problem 1 from Exercise Set 43 8

9 45 Summary of curve sketching We can now use many of the tools we have studied so far in Calculus, especially the table of monotonicity, the table of concavity and the asymptotes to develop a systematic and effective method for sketching the graph of any function f ( There are many important reasons for sketching the graph of a particular function by hand versus graphing it by calculator: even the best graphing devices have to be used intelligently and the graphs produced have to be fully understood; it is very important when graphing a given function to choose an appropriate viewing rectangle to avoid getting a misleading graph (see for example the graphs of f ( x 3,, f sin150x f ( x 3 150x f ( x 8, ( and other examples studied in section 14) Notions of Calculus (and algebra) are necessary to determine the domains and ranges of these functions (and therefore the appropriate drawing rectangles); using Calculus we discover the most important characteristics of a given function (such as asymptotes, minima, maxima and points of inflection) which can be easily overlooked or at best very hard to find accurately using only a graphical device For example, 3 Figure 1 shows the graph of f ( 8x 1x 18x It appears quite reasonable, it is the graph of a cubic, similar with the graph of 3 f ( x, apparently with no maximum or minimum value over this interval However, with the use of Calculus (the first derivative test), we find a maximum of f ( at x 075and a minimum at x 1 Calculus also insures that there are no other points of extrema for this function (if D R ) 3 Figure 1: The graph of f ( 8x 1x 18x Figure : The graph of Figure 1 zoomed in Therefore, there should be a close interaction between Calculus and graphing devices, as each can be of assistance to other: Calculus can provide the rigorous and accurate framework for finding the domain, range, asymptotes and all points of extrema of a given function, while a calculator (or another graphing device) can be used for carrying on necessary calculation or for checking our work) 9

10 Guideline (steps) for sketching the graph of a generic function f ( : 1 Find the domain D of f ( (the set of values x where f ( is well defined ) by imposing conditions for expressions in x ; Find Ox: by solving f ( 0 Find Oy: by calculating f (0) ; 3 (often optional) check for symmetries: with the y axis : f ( is symmetric with the y axis if f ( f ( f (, xd with origin: f ( is symmetric with the origin if ( check if f ( is periodic : is even, that is if f is odd, that is if f ( f (, x D f ( is periodic if pr such that f x p f (, x D The smallest such p is called the fundamental period of f ( If f ( is periodic with a fundamental period of T, then it needs to be graphed only on an interval of length T, typically on 0,T 4 Find the asymptotes: a) HA : at (if in D): Calculate lim f ( x at (if in D): Calculate lim f ( x b) VA are lines x a (points excluded from the domain D) such that either lim f ( (or ) lim f ( (or ) (or both) xa xa or 5 Draw the monotonicity and the concavity table of f ( (that is, determine the signs of f '( and f ''( on D ) and use the information from 1-4 above to complete this table 6 Sketch the graph of f ( using the information from the table of monotonicity and concavity drawn in 5 Example 1: Use the steps outlined above to sketch the graphs of: x 3 x a) f ( b) f ( x 3x f ( d) x 1 x 1 f ( x e x 47 Optimization problems : 10

11 In this section we solve practical problems in which we use the methods studied in sections 41 and 43 to find the extreme values of a practical quantity of interest There are many problems of this type, for example: a business person who wants to minimize costs and to maximize profit, a traveler wants to minimize his/her traveling time, or practical problems which come from physical or mathematical principles (such as the problem which generates the geodesic curves (the shortest curve between two points on a given surface), Fermat s principle of light which states that light follows the path which takes the least time See also the My dog knows calculus problem (discussed here: and mentioned in section 11) In this section we solve problems such as these, and also problems in which we are interested in maximizing areas, volumes, and profits and in minimizing distances, times and costs The problems in this section are typically word problems, somewhat similar with the related rates problems, studied in section 39, but with the notable difference that this time our main goal is to find the extreme values of a practical quantity of interest (usually called the objective function), and not a derivative It is important to follow the following guidelines when solving the optimization problems in this section: 1 Understand the problem: read the problem carefully until it is clearly understood Determine the objective function and the variables that this function depends on Draw a diagram: In most problems it is useful to draw a diagram to determine the given and the required quantities and the relation between these 3 Introduce notation: Assign a symbol for the objective function (let us call it Q ) and for other unknown quantities in the diagram (for example, call these a, b, c, x, y,) Label the diagram with these symbols 4 Express Q in terms of these symbols: a, b, c, x, y, Our final preliminary goal is to express Q in terms of one unknown variable only 5 Use the diagram (or equations/relations stated in the problem) to find relations between the unknown independent variables such that Q depends on one independent (unknown) variable only Set then Q f ( Find the domain of this function D 6 Use the methods studied in sections 41 and 43 (usually the First Derivative Test) to find the absolute minimum or the absolute maximum (as required) of Q f ( on D Examples : 1 A farmer has 400 feet of fence and he wants to fence off a rectangular field What are the dimensions of the field which has the largest area? A cylindrical can is to be made to hold 1000 cm 3 (1L) of oil, as in the Figure 3 below 11

12 What are the dimensions (find r and h ) which will minimize the cost of the metal to manufacture the can? 3 Find the point on the parabola y x which is closest to the point ( 1,4) Hint: it is easier here to minimize d x, y,(1,4) instead of x, y d,(1,4) 1

13 4 See and solve the My dog knows calculus problem at 13