4.4. Concavity and Curve Sketching. Concavity

Transcription

1 4.4 Concavit and Curve Sketching Concavit and Curve Sketching f' decreases CONCAVE DOWN 3 f' increases 0 CONCAVE UP FIGURE 4.25 The graph of ƒsd = 3 is concave down on s - q, 0d and concave up on s0, q d (Eample a). In Section 4.3 we saw how the first derivative tells us where a function is increasing and where it is decreasing. At a critical point of a differentiable function, the First Derivative Test tells us whether there is a local maimum or a local minimum, or whether the graph just continues to rise or fall there. In this section we see how the second derivative gives information about the wa the graph of a differentiable function bends or turns. This additional information enables us to capture ke aspects of the behavior of a function and its graph, and then present these features in a sketch of the graph. Concavit As ou can see in Figure 4.25, the curve = 3 rises as increases, but the portions defined on the intervals s - q, 0d and s0, q d turn in different was. As we approach the origin from the left along the curve, the curve turns to our right and falls below its tangents. The slopes of the tangents are decreasing on the interval s - q, 0d. As we move awa from the origin along the curve to the right, the curve turns to our left and rises above its tangents. The slopes of the tangents are increasing on the interval s0, q d. This turning or bending behavior defines the concavit of the curve.

2 268 Chapter 4: Applications of Derivatives DEFINITION Concave Up, Concave Down The graph of a differentiable function = ƒsd is (a) concave up on an open interval I if ƒ is increasing on I (b) concave down on an open interval I if ƒ is decreasing on I. If = ƒsd has a second derivative, we can appl Corollar 3 of the Mean Value Theorem to conclude that ƒ increases if ƒ 70 on I, and decreases if ƒ 60. The Second Derivative Test for Concavit Let = ƒsd be twice-differentiable on an interval I.. If ƒ 70 on I, the graph of ƒ over I is concave up. 2. If ƒ 60 on I, the graph of ƒ over I is concave down. CONCAVE UP '' 0 '' CONCAVE UP If = ƒsd is twice-differentiable, we will use the notations ƒ and interchangeabl when denoting the second derivative. EXAMPLE Appling the Concavit Test (a) The curve = 3 (Figure 4.25) is concave down on s - q, 0d where =6 6 0 and concave up on s0, q d where = (b) The curve = 2 (Figure 4.26) is concave up on s - q, q d because its second derivative =2 is alwas positive. EXAMPLE 2 Determining Concavit Determine the concavit of = 3 sin on [0, 2p]. FIGURE 4.26 The graph of ƒsd = 2 is concave up on ever interval (Eample b) sin '' sin 2 FIGURE 4.27 Using the graph of to determine the concavit of (Eample 2). Solution The graph of = 3 sin is concave down on s0, pd, where =-sin is negative. It is concave up on sp, 2pd, where =-sin is positive (Figure 4.27). Points of Inflection The curve = 3 sin in Eample 2 changes concavit at the point sp, 3d. We call sp, 3d a point of inflection of the curve. DEFINITION Point of Inflection A point where the graph of a function has a tangent line and where the concavit changes is a point of inflection. A point on a curve where is positive on one side and negative on the other is a point of inflection. At such a point, is either zero (because derivatives have the Intermediate Value Propert) or undefined. If is a twice-differentiable function, =0 at a point of inflection and has a local maimum or minimum.

3 4.4 Concavit and Curve Sketching 269 '' FIGURE 4.28 The graph of = 4 has no inflection point at the origin, even though =0 there (Eample 3). '' does not eist. 0 /3 FIGURE 4.29 A point where fails to eist can be a point of inflection (Eample 4). EXAMPLE 3 An Inflection Point Ma Not Eist Where =0 The curve = 4 has no inflection point at = 0 (Figure 4.28). Even though =2 2 is zero there, it does not change sign. EXAMPLE 4 An Inflection Point Ma Occur Where Does Not Eist The curve = >3 has a point of inflection at = 0 (Figure 4.29), but does not eist there. We see from Eample 3 that a zero second derivative does not alwas produce a point of inflection. From Eample 4, we see that inflection points can also occur where there is no second derivative. To stud the motion of a bod moving along a line as a function of time, we often are interested in knowing when the bod s acceleration, given b the second derivative, is positive or negative. The points of inflection on the graph of the bod s position function reveal where the acceleration changes sign. EXAMPLE 5 = d 2 d 2 a>3 b = d d a 3-2>3 b = >3. Studing Motion Along a Line A particle is moving along a horizontal line with position function sstd = 2t 3-4t 2 22t - 5, t Ú 0. Find the velocit and acceleration, and describe the motion of the particle. Solution The velocit is and the acceleration is std = s std = 6t 2-28t 22 = 2st - ds3t - d, astd = std = s std = 2t - 28 = 4s3t - 7d. When the function s(t) is increasing, the particle is moving to the right; when s(t) is decreasing, the particle is moving to the left. Notice that the first derivative s = s d is zero when t = and t = >3. Intervals Sign of Y s œ Behavior of s 0 6 t 6 increasing 6 t 6 >3 - decreasing >3 6 t increasing Particle motion right left right The particle is moving to the right in the time intervals [0, ) and s>3, q d, and moving to the left in (, >3). It is momentaril stationar (at rest), at t = and t = >3. The acceleration astd = s std = 4s3t - 7d is zero when t = 7>3. Intervals 0 6 t 6 7>3 7>3 6 t Sign of a s fl - Graph of s concave down concave up

4 270 Chapter 4: Applications of Derivatives The accelerating force is directed toward the left during the time interval [0, 7>3], is momentaril zero at t = 7>3, and is directed toward the right thereafter. Second Derivative Test for Local Etrema Instead of looking for sign changes in ƒ at critical points, we can sometimes use the following test to determine the presence and character of local etrema. THEOREM 5 Second Derivative Test for Local Etrema Suppose ƒ is continuous on an open interval that contains = c.. If ƒ scd = 0 and ƒ scd 6 0, then ƒ has a local maimum at = c. 2. If ƒ scd = 0 and ƒ scd 7 0, then ƒ has a local minimum at = c. 3. If ƒ scd = 0 and ƒ scd = 0, then the test fails. The function ƒ ma have a local maimum, a local minimum, or neither. f' 0, f'' 0 local ma f' 0, f'' 0 local min Proof Part (). If ƒ scd 6 0, then ƒ sd 6 0 on some open interval I containing the point c, since ƒ is continuous. Therefore, ƒ is decreasing on I. Since ƒ scd = 0, the sign of ƒ changes from positive to negative at c so ƒ has a local maimum at c b the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions = 4, = - 4, and = 3. For each function, the first and second derivatives are zero at = 0. Yet the function = 4 has a local minimum there, = - 4 has a local maimum, and = 3 is increasing in an open interval containing = 0 (having neither a maimum nor a minimum there). Thus the test fails. This test requires us to know ƒ onl at c itself and not in an interval about c. This makes the test eas to appl. That s the good news. The bad news is that the test is inconclusive if ƒ =0 or if ƒ does not eist at = c. When this happens, use the First Derivative Test for local etreme values. Together ƒ and ƒ tell us the shape of the function s graph, that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is turning or bending as defined b its concavit. We use this information to sketch a graph of the function that captures its ke features. EXAMPLE 6 Using ƒ and ƒ to Graph ƒ Sketch a graph of the function ƒsd = using the following steps. (a) Identif where the etrema of ƒ occur. (b) Find the intervals on which ƒ is increasing and the intervals on which ƒ is decreasing. (c) Find where the graph of ƒ is concave up and where it is concave down. (d) Sketch the general shape of the graph for ƒ.

5 4.4 Concavit and Curve Sketching 27 (e) Plot some specific points, such as local maimum and minimum points, points of inflection, and intercepts. Then sketch the curve. Solution ƒ is continuous since ƒ sd = eists. The domain of ƒ is s - q, q d, and the domain of ƒ is also s - q, q d. Thus, the critical points of ƒ occur onl at the zeros of ƒ. Since ƒ sd = = 4 2 s - 3d the first derivative is zero at = 0 and = 3. Intervals Sign of ƒ œ - - Behavior of ƒ decreasing decreasing increasing (a) Using the First Derivative Test for local etrema and the table above, we see that there is no etremum at = 0 and a local minimum at = 3. (b) Using the table above, we see that ƒ is decreasing on s - q, 0] and [0, 3], and increasing on [3, q d. (c) ƒ sd = = 2s - 2d is zero at = 0 and = 2. Intervals Sign of ƒ œ - Behavior of ƒ concave up concave down concave up We see that ƒ is concave up on the intervals s - q, 0d and s2, q d, and concave down on (0, 2). (d) Summarizing the information in the two tables above, we obtain <0 0<<2 2<<3 3< decreasing decreasing decreasing increasing concave up concave down concave up concave up The general shape of the curve is decr decr decr incr General shape. conc up conc down conc up conc up infl point infl point local min

6 272 Chapter 4: Applications of Derivatives 20 5 Inflection 0 point 5 (0, 0) Inflection (2, 6) 0 point (3, 7) Local minimum FIGURE 4.30 The graph of ƒsd = (Eample 6). (e) Plot the curve s intercepts (if possible) and the points where and are zero. Indicate an local etreme values and inflection points. Use the general shape as a guide to sketch the curve. (Plot additional points as needed.) Figure 4.30 shows the graph of ƒ. The steps in Eample 6 help in giving a procedure for graphing to capture the ke features of a function and its graph. Strateg for Graphing ƒ(). Identif the domain of ƒ and an smmetries the curve ma have. 2. Find and. 3. Find the critical points of ƒ, and identif the function s behavior at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if an occur, and determine the concavit of the curve. 6. Identif an asmptotes. 7. Plot ke points, such as the intercepts and the points found in Steps 3 5, and sketch the curve. EXAMPLE 7 Using the Graphing Strateg Sketch the graph of ƒsd = s d2 2. Solution. The domain of ƒ is s - q, q d and there are no smmetries about either ais or the origin (Section.4). 2. Find ƒ and ƒ. ƒsd = s d2 2 ƒ sd = s 2 d # 2s d - s d 2 # 2 s 2 d 2 = 2s - 2 d s 2 d 2 ƒ sd = s 2 d 2 # 2s -2d - 2s - 2 d[2s 2 d # 2] s 2 d 4 = 4s2-3d s 2 d 3 -intercept at = -, -intercept s = d at = 0 Critical points: = -, = After some algebra 3. Behavior at critical points. The critical points occur onl at = ; where ƒ sd = 0 (Step 2) since ƒ eists everwhere over the domain of ƒ. At = -, ƒ (-) = 7 0 ielding a relative minimum b the Second Derivative Test. At =, ƒ sd = ielding a relative maimum b the Second Derivative Test. We will see in Step 6 that both are absolute etrema as well.

7 4.4 Concavit and Curve Sketching Increasing and decreasing. We see that on the interval s - q, -d the derivative ƒ sd 6 0, and the curve is decreasing. On the interval s -, d, ƒ sd 7 0 and the curve is increasing; it is decreasing on s, q d where ƒ sd 6 0 again. 5. Inflection points. Notice that the denominator of the second derivative (Step 2) is alwas positive. The second derivative ƒ is zero when = -23, 0, and 23. The second derivative changes sign at each of these points: negative on A - q, - 23B, positive on A - 23, 0B, negative on A0, 23B, and positive again on A 23, q B. Thus each point is a point of inflection. The curve is concave down on the interval A - q, - 23B, concave up on A - 23, 0B, concave down on A0, 23B, and concave up again on A 23, q B. 6. Asmptotes. Epanding the numerator of ƒ() and then dividing both numerator and denominator b gives 2 2 (, 2) Point of inflection where 3 ƒsd = s d2 2 = = s2>d s>2 d s> 2. d Epanding numerator Dividing b 2 Point of inflection where 3 FIGURE 4.3 (Eample 7). Horizontal asmptote s d2 The graph of = 2 We see that ƒsd : as : q and that ƒsd : - as : - q. Thus, the line = is a horizontal asmptote. Since ƒ decreases on s - q, -d and then increases on s -, d, we know that ƒs -d = 0 is a local minimum. Although ƒ decreases on s, q d, it never crosses the horizontal asmptote = on that interval (it approaches the asmptote from above). So the graph never becomes negative, and ƒs -d = 0 is an absolute minimum as well. Likewise, ƒsd = 2 is an absolute maimum because the graph never crosses the asmptote = on the interval s - q, -d, approaching it from below. Therefore, there are no vertical asmptotes (the range of ƒ is 0 2). 7. The graph of ƒ is sketched in Figure 4.3. Notice how the graph is concave down as it approaches the horizontal asmptote = as : - q, and concave up in its approach to = as : q. Learning About Functions from Derivatives As we saw in Eamples 6 and 7, we can learn almost everthing we need to know about a twice-differentiable function = ƒsd b eamining its first derivative. We can find where the function s graph rises and falls and where an local etrema are assumed. We can differentiate to learn how the graph bends as it passes over the intervals of rise and fall. We can determine the shape of the function s graph. Information we cannot get from the derivative is how to place the graph in the -plane. But, as we discovered in Section 4.2, the onl additional information we need to position the graph is the value of ƒ at one point. The derivative does not give us information about the asmptotes, which are found using limits (Sections 2.4 and 2.5).

8 274 Chapter 4: Applications of Derivatives f() f() f () Differentiable smooth, connected; graph ma rise and fall ' 0 rises from left to right; ma be wav ' 0 falls from left to right; ma be wav or or '' 0 concave up throughout; no waves; graph ma rise or fall '' 0 concave down throughout; no waves; graph ma rise or fall '' changes sign Inflection point or ' changes sign graph has local maimum or local minimum ' 0 and '' 0 at a point; graph has local maimum ' 0 and '' 0 at a point; graph has local minimum