smi9885_ch3b.qd 5/7/ : PM Page 7 7 Chapter 3 Applications of Differentiation Debt loan, our indebtedness will decrease over time. If ou plotted our debt against time, the graph might look something like Figure 3.3. We now carefull define these notions. Notice that the following definition is merel a formal statement of something ou alread understand. Figure 3.3 Decreasing debt. Time Definition. A function f is (strictl) increasing on an interval I if for ever, I with <, f ( )< f ( ) [i.e., f () gets larger as gets larger]. A function f is (strictl) decreasing on the interval I if for ever, I with <, f ( )> f ( ) [i.e., f () gets smaller as gets larger]. Wh do we bother with such an obvious definition? Of course, anone can look at a graph of a function and immediatel see where that function is increasing and decreasing. The real challenge is to determine where a function is increasing and decreasing, given onl a mathematical formula for the function. For eample, can ou determine where f () = sin is increasing and decreasing, without looking at a graph? Even with a graph, can ou determine where this happens precisel? Look carefull at Figure 3.35 to see if ou can notice what happens at ever point at which the function is increasing or decreasing. f increasing (tangent lines have positive slope) f() f decreasing (tangent lines have negative slope) Figure 3.35 Increasing and decreasing. You should observe that on intervals where f is increasing, the tangent lines all have positive slope, while on intervals where f is decreasing, the tangent lines all have negative slope. Of course, we alread know that the slope of the tangent line at a point is given b the value of the derivative at that point. So, whether a function is increasing or decreasing on an interval seems to be connected to the sign of its derivative on that interval. This conjecture, although it s based onl on a single picture, sounds like a theorem and it is. Theorem. Suppose that f is differentiable on an interval I. (i) If f () > for all I, then f is increasing on I. (ii) If f () < for all I, then f is decreasing on I.
smi9885_ch3b.qd 5/7/ : PM Page 7 Section 3. Increasing and Decreasing Functions 7 Proof (i) Pick an two points, I, with <. Appling the Mean Value Theorem (Theorem 9. in section.9) to f on the interval (, ), we get f ( ) f ( ) = f (c), (.) for some number c (, ). (Wh can we appl the Mean Value Theorem here?) B hpothesis, f (c) > and since < (so that > ), we have from (.) that < f ( ) f ( ) or f ( )< f ( ). (.) Since (.) holds for all <, f is increasing on I. The proof of (ii) is nearl identical and is left as an eercise. What You See Ma Not Be What You Get One aim in the net few sections is to learn how to draw fairl representative graphs of functions (i.e., graphs that displa all of the significant features of a function: where it is increasing or decreasing, an etrema, asmptotes and two features we ll introduce in section 3.5: concavit and inflection points). You might think that there is little to be concerned about, given the ease with which ou can draw graphs b machine, but there is a significant issue here. When we draw a graph, we are drawing in a particular viewing window (i.e., a particular range of - and -values). When we use a computer or calculator to draw graphs, the window is often chosen b the machine. So, when is the window properl adjusted to show enough of the graph to be representative of the behavior of the function? We need to know when significant features are hidden outside of a given window and how to determine the precise locations of features that we can see in a given window. As we ll see, the onl wa we can resolve these questions is with a health dose of calculus. Eample. Drawing a Graph Draw a graph of f () = 3 + 9 showing all local etrema. Figure 3.36 = 3 + 9. Solution The most popular graphing calculators use the window defined b and as their default. Using this window, the graph of = f () looks like that displaed in Figure 3.36. The three segments seen in Figure 3.36 are not particularl revealing. Instead of blindl manipulating the window in the hope that a reasonable graph will magicall appear, we stop briefl to determine where the function is increasing and decreasing. First, we need the derivative f () = 6 + 8 = 6( + 3 ) = 6( )( + ). Note that the critical numbers are and, so these are the onl candidates for local etrema. We can see where the two factors and consequentl the derivative are positive
smi9885_ch3b.qd 5/7/ : PM Page 7 7 Chapter 3 Applications of Differentiation 8 5 Figure 3.37a = 3 + 9. ( ) 6( ) ( )( ) and negative from the number lines displaed in the margin. From this, note that f () >on(, ) (, ) f increasing. and f () <on(, ). f decreasing. So that ou can convenientl see this information at a glance, we have placed arrows indicating where the function is increasing and decreasing beneath the last number line. In particular, notice that this suggests that we are not seeing enough of the graph in the window in Figure 3.36. In Figure 3.37a, we redraw the graph in the window defined b 8 and 5 5. Here, we have selected the -range so that the critical points (, ) and (, 3) (and ever point in between) are displaed. Since we know that f is increasing on all of (, ), we know that the function is still increasing to the left of the portion displaed in Figure 3.37a. Likewise, since we know that f is increasing on all of (, ), we know that the function continues to increase to the right of the displaed portion. The graph in Figure 3.37a conves all of this significant behavior of the function. In Figure 3.37b, we have plotted both = f () (shown in blue) and = f () (shown in red). Notice the connection between the two graphs. When f () >, f is increasing; when f () <, f is decreasing and also notice what happens to f () at the local etrema of f. (We ll sa more about this shortl.) 8 5 Figure 3.37b = f () and = f (). You ma be tempted to think that ou can dispense with using an calculus to draw a graph. After all, it s eas to get lured into thinking that our graphing calculator or computer can draw a much better graph than ou can in onl a fraction of the time and with virtuall no effort. To some etent, this is true. You can draw graphs b machine and with a little fiddling with the graphing window, get a reasonable looking graph. Unfortunatel, this frequentl isn t enough. For instance, while it s clear that the graph in Figure 3.36 is incomplete, the initial graph in the following eample has a familiar shape and ma look reasonable, but it is incorrect. The calculus tells ou what features ou should epect to see in a graph. Without it, ou re simpl fooling around and hoping ou get something reasonable. Eample. Uncovering Hidden Behavior in a Graph Graph f () = 3 + 3.6. showing all local etrema....... f() Solution We first show the default graph drawn b our computer algebra sstem (see Figure 3.38a). We show a common default graphing calculator graph in Figure 3.38b. You can certainl make Figure 3.38b look more like Figure 3.38a b fooling around with the window some. But with some calculus, ou can discover features of the graph that would otherwise remain hidden. First, notice that f () = 3 +.. = (.)( + ) = (.)( +.)( + ). We show number lines for the three factors in the margin. Observe that { > on(,.) (., ) f f increasing. () < on(, ) (.,.). f decreasing.
smi9885_ch3b.qd 5/7/ : PM Page 73 Section 3. Increasing and Decreasing Functions 73 6 3 3 Figure 3.38a Default CAS graph of = 3 + 3.6.. Figure 3.38b Default calculator graph of = 3 + 3.6.. 5,, Figure 3.39a The global behavior of f () = 3 + 3.6.. 5 This sas that neither of the machine-generated graphs seen in Figures 3.38a or 3.38b are adequate, as the behavior on (, ) (.,.) cannot be seen in either graph. As it turns out, no single graph captures all of the behavior of this function. However, b increasing the range of -values to the interval [ 5, 5], we get the graph seen in Figure 3.39a. This shows the big picture, what we refer to as the global behavior of the function. Here, ou can see the local minimum at =, which was missing in our earlier graphs, but the behavior for values of close to zero is not clear. To see this, we need a separate graph, restricted to a smaller range of -values, as seen in Figure 3.39b. Notice that here, we can see the behavior of the function for close to zero quite clearl. In particular, the local maimum at =. and the local minimum at =. are clearl visible. We often sa that a graph such as Figure 3.39b shows the local behavior of the function. In Figures 3.a and 3.b, we show graphs indicating the global and local behavior of f () (in blue) and f () (in red) on the same set of aes. Pa particular attention to the behavior of f () in the vicinit of local etrema of f ()...,.3.3.3.3 5 5,.. Figure 3.39b Local behavior of f () = 3 + 3.6.. Figure 3.a = f () and = f () (global behavior). Figure 3.b = f () and = f () (local behavior). You ma have alread noticed a connection between local etrema and the intervals on which a function is increasing and decreasing. We state this in the following theorem.
smi9885_ch3b.qd 5/7/ : PM Page 7 7 Chapter 3 Applications of Differentiation f() f increasing Local maimum f() f decreasing Theorem. (First Derivative Test) Suppose that f is continuous on the interval [a, b] and c (a, b) is a critical number. (i) If f () > for all (a, c) and f () < for all (c, b) (i.e., f changes from increasing to decreasing at c), then f (c) is a local maimum. (ii) If f () < for all (a, c) and f () > for all (c, b) (i.e., f changes from decreasing to increasing at c), then f (c) is a local minimum. (iii) If f () has the same sign on (a, c) and (c, b), then f (c) is not a local etremum. c Figure 3.a Local maimum. It s easiest to think of this result graphicall. If f is increasing to the left of a critical number and decreasing to the right, then there must be a local maimum at the critical number (see Figure 3.a). Likewise, if f is decreasing to the left of a critical number and increasing to the right, then there must be a local minimum at the critical number (see Figure 3.b). B the wa, the preceding argument suggests a proof of the theorem. The job of writing out all of the details is left as an eercise. Eample.3 Finding Local Etrema Using the First Derivative Test c Find the local etrema of the function from eample., f () = 3 + 9. f() f decreasing Local minimum Figure 3.b Local minimum. f() f increasing Solution We had found in eample. that { >on(, ) (, ) f () <on(, ). f increasing. f decreasing. It now follows from the first derivative test that f has a local maimum located at = and a local minimum located at =. Theorem. works equall well for a function with critical points where the derivative is undefined. Eample. Finding Local Etrema of a Function with Fractional Eponents Find the local etrema of f () = 5/3 3 /3. Solution We have f () = 5 3 /3 3 = 5 6 3 /3, ( ) /3 3 6/5 (5 6) 3 /3 so that the critical numbers are 6 5 [ f ( 6 5 ) = ] and[f () is undefined]. Again drawing number lines for the factors, we determine where f is increasing and decreasing. Here, we have placed an above the on the number line for f () to indicate that f () is not defined at =. From this, we can see at a glance where f is positive and negative: 6/5 f() f () { >, on (, ) ( 6 5, ) <, on (, 6 5 ). f increasing. f decreasing.
smi9885_ch3b.qd 5/7/ : PM Page 75 Section 3. Increasing and Decreasing Functions 75 Consequentl, f has a local maimum at = and a local minimum at = 6 5. These local etrema are both clearl visible in the graph in Figure 3.. Figure 3. = 5/3 3 /3. Eample.5 Finding Local Etrema Approimatel Find the local etrema of f () = + 3 5 3 + 9 and draw a graph. Solution A graph of = f () using the most common graphing calculator default window appears in Figure 3.3. Without further analsis, we do not know if this graph shows all of the significant behavior of the function. [Note that some fourth-degree polnomials (e.g., f () = ) have graphs that look ver much like the one in Figure 3.3.] First, we compute Figure 3.3 f () = + 3 5 3 + 9. f () = 3 + 3. Unlike the last several eamples, this derivative does not easil factor. A graph of = f () (see Figure 3.) reveals three zeros, one near each of = 3,.5, and.5. Since a cubic polnomial has at most three zeros, there are no others. Using Newton s method or some other root-finding method [applied to f ()], we can find approimations to the three zeros of f. We get a.968, b.6386 and c.598. From Figure 3., we can see that f () is positive for a < < b and for > c and is negative elsewhere. That is, and f () >on(a, b) (c, ) f () <on(, a) (b, c). f increasing. f decreasing. You can quickl read off the local etrema: a local minimum at a.968, a local maimum at b.6386 and a local minimum at c.598. Since onl the local minimum at = c is visible in the graph in Figure 3.3, this graph is clearl not representative of the behavior of the function. B narrowing the range of displaed -values and widening the range of displaed -values, we obtain the far more useful graph seen in Figure 3.5. You should convince ourself, using the preceding analsis, that the local minimum at = c.598 is also the absolute minimum. 5 a b c a b c 5 5 Figure 3. f () = 3 + 3. Figure 3.5 f () = + 3 5 3 + 9.
smi9885_ch3b.qd 5/7/ : PM Page 76 76 Chapter 3 Applications of Differentiation EXERCISES 3.. Suppose that f () = and f () is an increasing function. To sketch the graph of = f (), ou could start b plotting the point (, ). Filling in the graph to the left, would ou move our pencil up or down? How does this fit with the definition of increasing?. Suppose ou travel east on an east-west interstate highwa. You reach our destination, sta a while and then return home. Eplain the First Derivative Test in terms of our velocities (positive and negative) on this trip. 3. Suppose that ou have a differentiable function f() with two critical numbers. Your computer has shown ou a graph that looks like the one below. In eercises 5 3, find the -coordinates of all etrema and sketch a graph. 5. = 3 + 6. = 3 + 7. = + + 8. = 5 + + 9. = +. = +. = e. = e 3. = ln. = e 5. = 6. = 7. = 3 8. = + 9. = sin + cos 3. = cos 3. = 3 + 3 3. = / / 33. = /3 /3 3. = /3 + /3 Discuss the possibilit that this is a representative graph: that is, is it possible that there are an important points not shown in this window?. Suppose that the function in eercise 3 has three critical numbers. Eplain wh the graph is not a representative graph. Eplain how ou would change the graphing window to show the rest of the graph. In eercises 5, find (b hand) the intervals where the function is increasing and decreasing. Verif our answers b graphing both f () and f (). 5. = 3 3 + 6. = 3 + + 7. = 8 + 8. = 3 3 9 + 9. = ( + ) /3. = ( ) /3. = sin 3. = sin 3. = e. = ln( ) In eercises 35, find the -coordinates of all etrema and sketch graphs showing global and local behavior of the function. 35. = 3 3 + 36. = 3 + 5 7 + 37. = 5 3 + 38. = 6 3. +.5 39. = 5 3 + 65. =.5 3. +. +. = ( + +.5)e. = 5 ln 8 In eercises 3 6, sketch a graph of a function with the given properties. 3. f () =, f () = 5, f () < for < and >, f () > for < <.
smi9885_ch3b.qd 5/7/ : PM Page 77 Section 3. Increasing and Decreasing Functions 77. f ( ) =, f () = 5, f () < for < and >, f () > for < <, f ( ) =, f () does not eist. 5. f (3) =, f () < for < and > 3, f () > for < < 3, f (3) =, f () and f () do not eist. 6. f () =, f () =, f () < for <, f () > for >, f () =. 7. Suppose an object has position function s(t), velocit function v(t) = s (t) and acceleration function a(t) = v (t). If a(t) >, then the velocit v(t) is increasing. Sketch two possible velocit functions, one with velocit getting less negative and one with velocit getting more positive. In both cases, sketch possible position graphs. The position graph should be curved (nonlinear). Does it look more like part of an upward-curving parabola or a downward-curving parabola? We look more closel at curving in section 3.5. 8. Repeat eercise 7 for the case where a(t) <. 9. Prove Theorem. (The First Derivative Test). 5. Give a graphical argument that if f (a) = g(a) and f () > g () for all > a, then f () >g() for all > a. Use the Mean Value Theorem to prove it. In eercises 5 5, use the result of eercise 5 to verif the inequalit. 5. > 3 for > 5. > sin for > 53. e > + for > 5. > ln for > 55. Give an eample showing that the following statement is false. If f () = and f () is a decreasing function, then the equation f () = has eactl one solution. 56. Determine if the following statement is true or false: If f () = and f () is an increasing function, then the equation f () = has no solutions. 57. Suppose that the total sales of a product after t months is given b s(t) = t + thousand dollars. Compute and interpret s (t). 58. In eercise 57, show that s (t) > for all t >. Eplain in business terms wh it is impossible to have s (t) <. 59. In this eercise, ou will pla the role of professor and construct a trick graphing eercise. The first goal is to find a function with local etrema so close together that the re difficult to see. For instance, suppose ou want local etrema at =. and =.. Eplain wh ou could start with f () = (.)( +.) =.. Look for a function whose derivative is as given. Graph our function to see if the etrema are hidden. Net, construct a polnomial of degree with two etrema ver near = and another near =. 6. In this eercise, we look at the abilit of fireflies to snchronize their flashes. (To see a remarkable demonstration of this abilit, see David Attenborough s video series Trials of Life.) Let the function f (t) represent an individual firefl s rhthm, so that the firefl flashes whenever f (t) equals an integer. Let e(t) represent the rhthm of a neighboring firefl, where again e(t) = n, for some integer n, whenever the neighbor flashes. One model of the interaction between fireflies is f (t) = ω + A sin[e(t) f(t)] for constants ω and A. If the fireflies are snchronized (e(t) = f (t)), then f (t) = ω, so the fireflies flash ever /ω time units. Assume that the difference between e(t) and f (t) is less than π. Show that if f (t) < e(t), then f (t) >ω. Eplain wh this means that the individual firefl is speeding up its flash to match its neighbor. Similarl, discuss what happens if f (t) > e(t). 6. The HIV virus attacks specialized T cells that trigger the human immune sstem response to a foreign substance. If T(t) is the population of uninfected T cells at time t (das) and V(t) is the population of infectious HIV in the bloodstream, a model that has been used to stud AIDS is given b the following differential equation that describes the rate at which the population of T cells changes. [ T (t) = + + V (t) ] T (t)v (t).t(t) +. + V (t).t (t)v(t). If there is no HIV present [that is, V(t) = ] and T(t) =, show that T (t) =. Eplain wh this means that the T-cell count will remain constant at (cells per cubic mm). Now, suppose that V(t) =. Show that if T (t) is small enough, then T (t) > and the T-cell population will increase. On the other hand, if T (t) is large enough, then T (t) < and the T-cell population will decrease. For what value of T(t) is T (t) =? Even though this population would remain stable, eplain wh this would be bad news for the infected human.
smi9885_ch3b.qd 5/7/ : PM Page 78 78 Chapter 3 Applications of Differentiation 3.5 CONCAVITY In section 3., we saw how to determine where a function is increasing and decreasing and how this relates to drawing a graph of the function. At this point, we need to see how to refine the shape of a graph. First, ou must realize that simpl knowing where a function increases and decreases is not sufficient to draw good graphs. In Figures 3.6a and 3.6b, we show two ver different shapes of increasing functions joining the same two points. a b a b Figure 3.6a Increasing function. Figure 3.6b Increasing function. So, given that a graph joins two particular points and is increasing, which of the two shapes shown do we draw? Without further information, there s no wa to tell. Realize that this is an important distinction to make. For eample, suppose that Figure 3.6a or 3.6b depicts the balance in our bank account. Both indicate a balance that is growing. Notice that the rate of growth in Figure 3.6a, is increasing, while the rate of growth depicted in Figure 3.6b is decreasing. Which would ou want to have describe our bank balance? Wh? Figures 3.7a and 3.7b are the same as Figures 3.6a and 3.6b, respectivel, but with a few tangent lines drawn in. a b a b Figure 3.7a Concave up increasing. Figure 3.7b Concave down increasing. Notice that although all of the tangent lines have positive slope [since f () > ], the slopes of the tangent lines in Figure 3.7a are increasing, while those in Figure 3.7b are decreasing. We call the graph in Figure 3.7a concave up and the graph in Figure 3.7b concave down. The situation is similar for decreasing functions. In Figures 3.8a and 3.8b, we show two different shapes of decreasing function. Again, although both functions are decreasing, the one shown in Figure 3.8a is concave up (slopes of tangent lines
smi9885_ch3b.qd 5/7/ : PM Page 79 Section 3.5 Concavit 79 increasing) and the one shown in Figure 3.8b is concave down (slopes of tangent lines decreasing). We have the following definition. Definition 5. For a function f that is differentiable on an interval I, the graph of f is (i) concave up on I, if f is increasing on I or (ii) concave down on I, if f is decreasing on I. a Figure 3.8a Concave up. b How can ou tell when f is increasing or decreasing? The derivative of f (i.e., f ) ields that information. The following theorem connects this definition with what we alread know about increasing and decreasing functions. The proof of the theorem is a straightforward application of Theorem. to Definition 5.. Theorem 5. Suppose that f eists on an interval I. (i) If f () > on I, then the graph of f is concave up on I. (ii) If f () < on I, then the graph of f is concave down on I. a b Figure 3.8b Concave down. Eample 5. Determining Concavit Determine where the graph of f () 3 + 9 is concave up and concave down and draw a graph showing all significant behavior of the function. Inflection point 5 Figure 3.9 = 3 + 9. N OTES If (c, f (c)) is an inflection point, then either f (c) or f (c) is undefined. So, finding all points where f () is zero or is undefined gives ou all possible candidates for inflection points. But beware: not all points where f () is zero or undefined correspond to inflection points. Solution Here, we have f () 6 + 8 and from our work in eample.3, we have { > on(, ) (, ) f () < on(, ). We now have f () = + 8 f increasing. f decreasing. { >, for > 3 <, for < 3. Concave up. Concave down. Using all of this information, we are able to draw the graph shown in Figure 3.9. Notice that at the point ( 3, f ( 3 )), the graph changes from concave down to concave up. Such points are called inflection points. More precisel, we have the following definition. Definition 5. Suppose that f is continuous on the interval (a, b) and that the graph changes concavit at a point c (a, b) (i.e., the graph is concave down on one side of c and concave up on the other). Then, the point (c, f (c)) is called an inflection point of f.
smi9885_ch3b.qd 5/7/ : PM Page 8 8 Chapter 3 Applications of Differentiation 3 3 3 3 3 3 f() ( ) ( ) f() Eample 5. Determine where the graph of f () 6 + is concave up and concave down, find an inflection points and draw a graph showing all significant features. Solution Here, we have f () = 3 = ( 3) = ( 3)( + 3). We have drawn number lines for the factors of f () in the margin. From this, notice that { >, on ( 3, ) ( 3, ) f () <, on (, 3) (, f increasing. 3). f decreasing. Net, we have Determining Concavit and Locating Inflection Points 3 3 f() f() f () = = ( )( + ). We have drawn number lines for the two factors in the margin. From this, we can see that { >, on (, ) (, ) f Concave up. () <, on (, ). Concave down. 3 5 5 3 So that ou can see this information at a glance, we have indicated the concavit below the bottom number line, with small concave up and concave down segments. Finall, observe that since the graph changes concavit at and, there are inflection points located at (, ) and (, ). Using all of this information, we are able to draw the graph shown in Figure 3.5. For our convenience, we have reproduced the number lines for f () and f () together above the figure. Figure 3.5 = 6 +. As we see in the following eample, having f () does not impl the eistence of an inflection point. Eample 5.3 A Graph with No Inflection Points Determine the concavit of f() and locate an inflection points. Solution There s nothing trick about this function. We have f () 3 and f (). Since f () > for > and f () < for <, we know that f is increasing for > and decreasing for <. Further, f () > for all, while f (). So, the graph is concave up for. Further, even though f (), there is no inflection point at. We show a graph of the function in Figure 3.5. Figure 3.5 =. There is also a connection between second derivatives and etrema. Suppose that f (c) and that the graph of f is concave down in some open interval containing c. Then, nearb c, the graph looks like that in Figure 3.5a and hence, f(c) is a local maimum.
smi9885_ch3b.qd 5/7/ : PM Page 8 Section 3.5 Concavit 8 f(c) Likewise, if f (c) and the graph of f is concave up in some open interval containing c, then nearb c, the graph looks like that in Figure 3.5b and hence, f (c) is a local minimum. More precisel, we have the following theorem. c f(c) Figure 3.5a Local maimum. Theorem 5. (Second Derivative Test) Suppose that f is continuous on the interval (a, b) and f (c), for some number c (a, b). (i) If f (c) <, then f (c) is a local maimum and (ii) if f (c) >, then f (c) is a local minimum. f(c) We leave a formal proof of this theorem as an eercise. When appling the theorem, simpl think about Figures 3.5a and 3.5b. Eample 5. Using the Second Derivative Test to Find Etrema c f(c) Use the second derivative test to find the local etrema of f () = 8 +. Solution Here, Figure 3.5b Local minimum. f () = 3 6 = ( ) = ( )( + ). Thus, the critical numbers are =, and. We can test these using the second derivative test as follows. We have and so, f () = 6 f () = 6 <, and f ( ) = 3 > f () = 3 >. Figure 3.53 = 8 +. Thus, f () is a local maimum and f ( ) and f () are local minima. We show a graph of = f () in Figure 3.53. Remark 5. If f (c) = or f (c) is undefined, the second derivative test ields no conclusion. That is, f (c) ma be a local maimum, a local minimum or neither. In this event, we must rel solel on first derivative information (i.e., the first derivative test) to determine if f (c) is a local etremum or not. We illustrate this with the following eample.
smi9885_ch3b.qd 5/7/ : PM Page 8 8 Chapter 3 Applications of Differentiation 3 Eample 5.5 Functions for Which the Second Derivative Test Is Inconclusive Use the second derivative test to tr to classif an local etrema for (a) f () = 3, (b) g() = and (c) h() =. 3 Figure 3.5a = 3. Solution (a) Note that f () = 3 and f () = 6. So, the onl critical number is = and f () =, also. We leave it as an eercise to show that the point (, ) is not a local etremum (see Figure 3.5a). (b) We have g () = 3 and g () =. Again, the onl critical number is = and g () =. In this case, though, g () < for < and g () > for >. So, b the first derivative test, (, ) is a local minimum (see Figure 3.5b). (c) Finall, we have h () = 3 and h () =. Once again, the onl critical number is =, h () = and we leave it as an eercise to show that (, ) is a local maimum for h (see Figure 3.5c). We can use first and second derivative information to help produce a meaningful graph of a function, as in the following eample. Figure 3.5b =. Eample 5.6 Drawing a Graph of a Rational Function Draw a graph of f () = + 5, showing all significant features. Solution The first thing that ou should notice here is that the domain of f consists of all real numbers other than =. Net, we have f () = 5 = 5 Add the fractions. = ( 5)( + 5). So, the onl two critical numbers are = 5, 5. (Wh is = not a critical number?) Looking at the three factors in f (), we get the number lines shown in the margin. Thus, { >, on (, 5) (5, ) f () <, on ( 5, ) (, 5). f increasing. f decreasing. Further, 5 5 Figure 3.5c =. 5 5 5 5 f() f () = 5 { >, on (, ) 3 <, on (, ). Concave up. Concave down. Be careful here. There is no inflection point on the graph, even though the graph is concave up on one side of = and concave down on the other. (Wh not?) We can now use either the first derivative test or the second derivative test to determine the local etrema. B the second derivative test, f (5) = 5 5 >
smi9885_ch3b.qd 5/7/ : PM Page 83 Section 3.5 Concavit 83 and f ( 5) = 5 5 <, 5 5 5 Figure 3.55 = + 5. 5 so that there is a local minimum at = 5 and a local maimum at = 5. Finall, before we can draw a representative graph, we need to know what happens to the graph near =, since is not in the domain of f. We have and + f () = + ( ( f () = + 5 + 5 ) = ) =. So, there is a vertical asmptote at =. Putting together all of this information, we get the graph shown in Figure 3.55. In eample 5.6, we computed f () and f () to uncover the behavior of the + function near =. We needed to consider this since = was not in the domain of f. In the following eample, we ll see that since = is not in the domain of f (although it is in the domain of f ), we must compute f () and f (). This will tell us + about the behavior of the tangent lines near =. Eample 5.7 A Function with a Vertical Tangent Line at an Inflection Point Draw a graph of f () = ( + ) /5 +, showing all significant features. Solution First, notice that the domain of f is the entire real line. We also have f () = 5 ( + ) /5 >, for. So, f is increasing everwhere, ecept at = [the onl critical number, where f ( ) is undefined]. This also sas that f has no local etrema. Further, 6 f () = 5 ( + ) 9/5 { >, on (, ) <, on (, ). Concave up. Concave down. 3 So, there is an inflection point at =. In this case, f () is undefined at =. Since is in the domain of f, but not in the domain of f, we consider f () = 5 ( + ) /5 = and f () = + + 5 ( + ) /5 =. Figure 3.56 = ( + ) /5 +. This sas that the graph has a vertical tangent line at =. We obtain the graph shown in Figure 3.56.
smi9885_ch3b.qd 5/7/ : PM Page 8 8 Chapter 3 Applications of Differentiation EXERCISES 3.5. It is often said that a graph is concave up if it holds water. This is certainl true for parabolas like =, but is it true for graphs like = /? It is alwas helpful to put a difficult concept into everda language, but the danger is in oversimplification. Do ou think that holds water is helpful or can it be confusing? Give our own description of concave up, using everda language. (Hint: One popular image involves smiles and frowns.) 6.. Find a reference book with the population of the United States since 8. From 8 to 9, the numerical increase b decade increased. Argue that this means that the population curve is concave up. From 96 to 99, the numerical increase b decade has been approimatel constant. Argue that this means that the population curve is near a point of zero concavit. Wh does this not necessaril mean that we are at an inflection point? Argue that we should hope, in order to avoid overpopulation, that it is indeed an inflection point. 7. 3. The goal of investing in the stock market is to bu low and sell high. But, how can ou tell whether a price has peaked or not? Once a stock price goes down, ou can see that it was at a peak but then it s too late to do anthing about it! Concavit can help. Suppose a stock price is increasing and the price curve is concave up. Wh would ou suspect that it will continue to rise? Is this a good time to bu? Now, suppose the price is increasing but the curve is concave down. Wh should ou be preparing to sell? Finall, suppose the price is decreasing. If the curve is concave up, should ou bu or sell? What if the curve is concave down?. Suppose that f (t) is the amount of mone in our bank account at time t. Eplain in terms of spending and saving what would cause f (t) to be decreasing and concave down; increasing and concave up; decreasing and concave up. In eercises 5 8, estimate the intervals where the function is concave up and concave down. (Hint: Estimate where the slope is increasing and decreasing.) 8. 3 5 5 5 5 3 5. In eercises 9, find the intervals of increase and decrease, all local etrema, the intervals of concavit, all inflection points and sketch a graph. 9. f () = 3 3 +. f () = 3 + 3 6. f () = +. f () = + 3 3 3. f () = + /. f () = 6/ 5. f () = 3 6 + 6. f () = 3 + 3
smi9885_ch3b.qd 5/7/ : PM Page 85 Section 3.5 Concavit 85 7. f () = + 3 8. f () = + + 9. f () = e. f () = e In eercises 7 and 8, estimate the intervals of increase and decrease, the locations of local etrema, intervals of concavit and locations of inflection points.. f () = 9. f () = 9 7. 3. f () = ( + ) /3. f () = ln 8 5. f () = 9 6. f () = + 6 7. f () = sin + cos 8. f () = + cos 9. f () = e sin 3. f () = e cos 3. f () = 3/ / 3. f () = /3 /3 3 3 33. f () = 3 3. f () = 3 + 35. f () = 6 3 + 8. 36. f () = 3 + 7 37. f () = 5 + 3 38. f () = 39. f () = 6 3 + 39.6 +. f () = + 3 3..8 In eercises 6, sketch a graph with the given properties.. f () =, f () > for all, f () =, f () > for >, f () < for <, f () =. f () =, f () for all, f () =, f () > for >, f () < for <, f () = 3. f () =, f () > for < and < <, f () < for >, f () > for <, < < and >, f () < for < <. f () =, f () > for all, f () =, f () > for <, f () < for > 5. f () =, f ( ) =, f () =, f () > for < and < <, f () < for < < and >, f () < for < and > 6. f () =, f () < for <, f () > for >, f () < for < and > 9. Repeat eercise 7 if the given graph is of f () instead of f (). 5. Repeat eercise 8 if the given graph is of f () instead of f (). 5. Suppose that w(t) is the depth of water in a cit s water reservoir. Which would be better news at time t =,w () =.5 or w () =.5 or would ou need to know the value of w () to determine which is better? 5. Suppose that T (t) is a sick person s temperature at time t. Which would be better news at time t, T () = or T () = or would ou need to know the value of T () and T () to determine which is better? 53. Suppose that a compan that spends $ thousand on advertising sells $s() of merchandise, where s() = 3 3 + 7 36 + 8,. Find the value of that maimizes the rate of change of sales. (Hint: Read the question carefull!) 5. For the sales function in eercise 53, find the inflection point and eplain wh in advertising terms this is the point of diminishing returns.
smi9885_ch3b.qd 5/7/ : PM Page 86 86 Chapter 3 Applications of Differentiation 55. Suppose that it costs a compan C() =. + + 36 dollars to manufacture units of a product. For this cost function, the average cost function is C() = C(). Find the value of that minimizes the average cost. 56. In eercise 55, the cost function can be related to the efficienc of the production process. Eplain wh a cost function that is concave down indicates better efficienc than a cost function that is concave up. 57. Show that there is an inflection point at (, ) for an function of the form f () = + c 3, where c is a nonzero constant. What role(s) does c pla in the graph of = f ()? 58. The following eamples show that there is not a perfect match between inflection points and places where f () =. First, for f () = 6, show that f () =, but there is no inflection point at =. Then, for g() =, show that there is an inflection point at =, but that g () does not eist. 59. Give an eample of a function showing that the following statement is false. If the graph of = f () is concave down for all, the equation f () = has at least one solution. 6. Determine if the following statement is true or false. If f () =, f () eists for all and the graph of = f () is concave down for all, the equation f () = has at least one solution. 6. One basic principle of phsics is that light follows the path of minimum time. Assuming that the speed of light in the earth s atmosphere decreases as altitude decreases, argue that the path that light follows is concave down. Eplain wh this means that the setting sun appears higher in the sk than it reall is. 6. Prove Theorem 5. (the Second Derivative Test). (Hint: Think about what the definition of f (c) sas when f (c) > or f (c) <.) 63. The linear approimation that we defined in section 3. is the line having the same location and the same slope as the function being approimated. Since two points determine a line, two requirements (point, slope) are all that a linear function can satisf. However, a quadratic function can satisf three requirements since three points determine a parabola (and there are three constants in a general quadratic function a + b + c). Suppose we want to define a quadratic approimation to f () at = a. Building on the linear approimation, the general form is g() = f (a) + f (a)( a) + c( a) for some constant c to be determined. In this wa, show that g(a) = f (a) and g (a) = f (a). That is, g() has the right position and slope at = a. The third requirement is that g() have the right concavit at = a, so that g (a) = f (a). Find the constant c that makes this true. Then, find such a quadratic approimation for each of the functions sin, cos and e at =. In each case, graph the original function, linear approimation and quadratic approimation and describe how close the approimations are to the original functions. 6. In this eercise, we eplore a basic problem in genetics. Suppose that a species reproduces according to the following probabilities: p is the probabilit of having no children, p is the probabilit of having one offspring, p is the probabilit of having two offspring,..., p n is the probabilit of having n offspring and n is the largest number of offspring possible. Eplain wh for each i, we have p i and p + p + p + + p n =. We define the function F() = p + p + p + + p n n. The smallest nonnegative solution of the equation F() = for represents the probabilit that the species becomes etinct. Show graphicall that if p > and F () >, then there is a solution of F() = with < <. Thus, there is a positive probabilit of survival. However, if p > and F () <, show that there are no solutions of F() = with < <. (Hint: First show that F is increasing and concave up.) 3.6 OVERVIEW OF CURVE SKETCHING You might be wondering wh ou need to spend an more time on curve sketching. We have alread drawn numerous graphs over the last three sections. Besides, with a graphing calculator or computer algebra sstem at our disposal, wh must ou even consider drawing graphs b hand? Of course, graphing calculators or computer algebra sstems are powerful tools toda in the stud or application of mathematics. As the authors of this tet, we admit it. We have
smi9885_ch3b.qd 5/7/ : PM Page 87 Section 3.6 Overview of Curve Sketching 87 6 8 3 3 Figure 3.57a = + 6 3 + + 8 + (one view). Figure 3.57b = + 6 3 + + 8 + (standard calculator view). made etensive use of several computer algebra sstems (Maple and Mathematica) in preparing the manuscript for this tet and even in designing our problems. So then, wh should we condemn ou to drawing graphs b hand? We re certain that late night talk show hosts would have a list of the top reasons wh sadistic mathematics professors would want to inflict such pain on their students, but there s reall onl one reason. For better or worse, graphing calculators and computer algebra sstems do not actuall draw graphs. What the do is plot points (albeit lots of them) and then connect the points with a smooth curve. Of course, this works eceptionall well for some functions, but leaves something to be desired for others. The problem boils down to the window in which ou draw the graph and how man points ou plot in that window. The onl wa to know how to choose the window or how man points to plot in that window is to use the calculus to determine the properties of the graph that ou are interested in seeing. We have alread made this point a number of times. The calculus tells us the properties that a representative graph should ehibit. We then tr to adjust the window and the number of points plotted in order to produce such a graph. We begin this section b summarizing the various tests that ou should perform on a function when tring to draw a graph of = f (). Domain: You should alwas determine the domain of f first. Vertical Asmptotes: For an isolated point not in the domain of f, check the iting value of the function as approaches that point, to see if there is a vertical asmptote or a jump or removable discontinuit at that point. First Derivative Information: Determine where f is increasing and decreasing and find an local etrema. Vertical Tangent Lines: At an isolated point not in the domain of f, but in the domain of f, check the iting values of f (), to determine if there is a vertical tangent line at that point. Second Derivative Information: Determine where the graph is concave up and concave down and locate an inflection points. Horizontal Asmptotes: Check the it of f () as and as. Intercepts: Locate - and -intercepts, if an. If this can t be done eactl, then do so approimatel (e.g., using Newton s method). We start with a ver simple eample. Eample 6. Drawing a Graph of a Polnomial Draw a graph of f () = + 6 3 + + 8 + showing all significant features. Solution Wh shouldn t we simpl ask our computer algebra sstem for a graph? Computer algebra sstems and graphing calculators usuall do one of two things to determine the window in which the will displa a graph. (Some give the user an option as to how to set the default method.) One method is to compute a set number of function values over a given standard range of -values. The -range is then chosen so that all of the calculated points can be displaed. This might result in a graph that looks like the one in Figure 3.57a, produced using the computer algebra sstem Maple. Another method is to draw a graph in a fied, default window. For instance, most graphing calculators use the default window defined b and. Using this window, we get the graph shown in Figure 3.57b. Of course, these two graphs are ver different. Without the calculus, it s difficult to tell which, if either, of these is
smi9885_ch3b.qd 5/7/ : PM Page 88 88 Chapter 3 Applications of Differentiation q 3 q 8 6 Figure 3.58 = + 6 3 + + 8 +. q ( ) ( ) f() ( ) ( ) f() f() f() trul representative of the behavior of f. Some analsis will help to clear up the situation. First, note that the domain of f is the entire real line. Further, since f () is a polnomial, it doesn t have an vertical or horizontal asmptotes. (Think about this!) Net, note that f () = 3 + 8 + + 8 = ( + )( + ). Drawing number lines for the individual factors in f (), we have that { ( >, on f (), ) <, on (, ) ( f increasing., ). f decreasing. This also tells us that there is a local minimum at = and that there are no local maima. Net, we have f () = + 36 + = ( + )( + ). Drawing number lines for the factors of f (), we have { >, on (, ) (, ) f Concave up. () <, on (, ). Concave down. From this, we see that there are inflection points at = and at =. Finall, to find the -intercepts, we need to solve f () = approimatel. Doing this (we leave the details as an eercise: use Newton s method or our calculator s solver), we find that there are two -intercepts: = (eactl) and.673. Notice that the significant -values that we have identified are =, = and =. Computing ( the corresponding -values from = f (), we get the points (, ), (, ) and, 6). We summarize the first and second derivative information in the number lines in the margin. In Figure 3.58, we include all of these important points b setting the -range to be 3 and the -range to be 8. In the following eample, we eamine a function that has local etrema, inflection points and both vertical and horizontal asmptotes. Eample 6. Drawing a Graph of a Rational Function 3e e e 3e Figure 3.59a = 3 3. Draw a graph of f () = 3 3 showing all significant features. Solution The default graph drawn b our computer algebra sstem (Maple) appears in Figure 3.59a. Notice that this doesn t seem to be a particularl useful graph, since ver little is visible (or at least distinguishable from the aes). The graph drawn using the most common graphing calculator default window (with a range of and ) is seen in Figure 3.59b. This is arguabl an improvement over Figure 3.59a, but does this graph conve all that it could about the function (e.g., about local etrema, inflection points, etc.)? We can answer this question onl after we do some calculus. We follow the outline given at the beginning of the section. First, observe that the domain of f includes all real numbers. Since = is an isolated point not in the domain of f, we scrutinize the iting behavior of f as
smi9885_ch3b.qd 5/7/ : PM Page 89 Section 3.6 Overview of Curve Sketching 89 3 3 8 8 Figure 3.59b = 3 3. 8 3 3 8 f() 3 3 f() 8 8 5 approaches. We have and + 3 f () = = (6.) + 3 + 3 f () = =. (6.) 3 From (6.) and (6.), we see that the graph has a vertical asmptote at =. Net, we look for whatever information the first derivative will ield. We have f () = (3 ) ( 3)(3 ) ( 3 ) Quotient rule. = [ 3( 3)] 6 Factor out an. = 9 Combine terms. (3 )(3 + ) =. Factor difference of two squares. Looking at the individual factors in f (), we have the number lines shown in the margin. Thus, { f >, on ( 3, ) (, 3) f increasing. () (6.3) <, on (, 3) (3, ). f decreasing. Note that this sas that f has a local minimum at = 3and a local maimum at = 3. Net, we look at f () = ( ) (9 )( 3 ) ( ) Quotient rule. = 3 [ + (9 )()] 8 Factor out 3. = (8 ) 5 Combine terms. = ( 8)( + 8) 5. Factor difference of two squares. Looking at the individual factors in f (), we obtain the number lines found in the margin. Thus, we have { >, on ( 8, ) ( 8, ) f () <, on (, 8) (, Concave up. (6.) 8). Concave down. This sas that there are inflection points at =± 8. (Wh is there no inflection point at =?) To determine the iting behavior as ±, we consider 3 f () = = 3 ( 3 3 ) =. (6.5)
smi9885_ch3b.qd 5/7/ : PM Page 9 9 Chapter 3 Applications of Differentiation 3 8 5.. 5 Figure 3.6 = 3 3. 3 8 f() f() Likewise, we have f () =. (6.6) So, the line = is a horizontal asmptote both as and as. Finall, the -intercepts are where = f () = 3, 3 that is, at =± 3. Notice that there are no -intercepts, since = is not in the domain of the function. We now have all of the information that we need to draw a representative graph. With some eperimentation, ou can set the - and -ranges so that most of the significant features of the graph (i.e., vertical and horizontal asmptotes, local etrema, inflection points, etc.) are displaed, as in Figure 3.6. Notice that the graph in Figure 3.6 is consistent with all of the information that we accumulated on the function in (6.) (6.6). Although the eistence of the inflection points is clearl indicated b the change in concavit, their precise location is as et a bit fuzz in this graph. Notice, however, that both vertical and horizontal asmptotes and the local etrema are clearl indicated, something which cannot be said about either of Figures 3.59a or 3.59b. In the following eample, there are multiple vertical asmptotes, onl one etremum and no inflection points. Eample 6.3 A Graph with Two Vertical Asmptotes 5 5 5 Draw a graph of f () = showing all significant features. Solution The default graph produced b our computer algebra sstem is seen in Figure 3.6a, while the default graph drawn b most graphing calculators looks like the graph seen in Figure 3.6b. Notice that the domain of f includes all ecept =± (since the denominator is zero at =±). Figure 3.6b suggests that there are vertical asmptotes at =±, but let s establish this carefull. We have Figure 3.6a =. + = + ( )( + ) + + + =. (6.7) Similarl, we get 5 5 5 5 and =, = (6.8) + =. (6.9) Figure 3.6b =. Thus, there are vertical asmptotes at =±. Net, we have f () = ( ) () = 8 ( ) ( ).
smi9885_ch3b.qd 5/7/ : PM Page 9 Section 3.6 Overview of Curve Sketching 9 6 6 6 Figure 3.6 =. 6 ( ) 3 ( ) 3 f() f() f() Since the denominator is positive for ±, it is a simple matter to see that { >, on (, ) (, ) f f increasing. () (6.) <, on (, ) (, ). f decreasing. In particular, notice that the onl critical number is = (since =, are not in the domain of f ). Thus, the onl local etremum is the local maimum located at =. Net, we have f () = 8( ) + (8)( ) () ( ) Quotient rule. = 8( )[ ( ) + ] ( ) Factor out 8( ). = 8(3 + ) ( ) 3 Combine terms. = 8(3 + ). ( ) 3 Factor difference of two squares. ( + ) 3 Since the numerator is positive for all, we need onl consider the terms in the denominator, as seen in the margin. We then have { f >, on (, ) (, ) Concave up. () (6.) <, on (, ). Concave down. However, since =, are not in the domain of f, there are no inflection points. It is an eas eercise to verif that and = (6.) =. (6.3) From (6.) and (6.3), we have that = is a horizontal asmptote, both as and as. Finall, we observe that the onl -intercept is at =. We can now summarize the information in (6.7) (6.3) in the graph seen in Figure 3.6. In the following eample, we need to use computer-generated graphs, as well as a rootfinding method to determine the behavior of the function. Eample 6. Graphing Where the Domain and Etrema Must be Approimated Draw a graph of f () = 3 + 3 showing all significant features. + 3 + 3 Figure 3.63 = 3 + 3 + 3 + 3. Solution The default graph drawn b most graphing calculators and computer algebra sstems looks something like the one shown in Figure 3.63. This seems to reveal more about the function than did its counterparts in eample 6.3, but we can onl determine all the significant features b doing some calculus. Since f is a rational function, it is defined for all, ecept for where the denominator is zero, that is, where 3 + 3 + 3 + 3 =.