Section 4.4 Concavity and Points of Inflection

Transcription

1 Section 4.4 Concavit and Points of Inflection In Chapter 3, ou saw that the second derivative of a function has applications in problems involving velocit and acceleration or in general rates-of-change problems. Here we eamine the use of the second derivative of a function in curve sketching. INVESTIGATION 1 The purpose of this investigation is to eamine the relationship between slopes of tangents and the second derivative of a function. A. Sketch the graph of f B. Determine f 12. Use f 12 to calculate the slope of the tangent to the curve at the points with the following -coordinates: 4, 3, 2, 1,,1,2,3, and 4. Sketch each of these tangents. C. Are these tangents above or below the graph of f 12? D. Describe the change in the slopes as increases. E. Determine f 12. How does the value of f 12 relate to the wa in which the curve opens? How does the value of f 12 relate to the wa f 12 changes as increases? F. Repeat parts B, C, and D for the graph of f G. How does the value of f 12 relate to the wa in which the curve opens? INVESTIGATION 2 The purpose of this investigation is to etend the results of Investigation 1 to other functions. A. Sketch the graph of f B. Determine all the values of for which f 12. C. Determine intervals on the domain of the function such that f 12 6, f 12, and f D. For values of such that f 12 6, how does the shape of the curve compare with our conclusions in Investigation 1? E. Repeat part D for values of such that f F. What happens when f 12? G. Using our observations from this investigation, sketch the graph of CONCAVITY AND POINTS OF INFLECTION

2 From these investigations, we can make a summar of the behaviour of the graphs. Concavit and the Second Derivative 1. The graph of f 12 is concave up on an interval a b in which the slopes of f 12 are increasing. On this interval, f 12 eists and f The graph of the function is above the tangent at ever point on the interval. a b a b 2. The graph of f 12 is concave down on an interval a b in which the slopes of f 12 are decreasing. On this interval, f 12 eists and f The graph of the function is below the tangent at ever point on the interval. a b a b 3. If f 12 has a critical point at c, with f 1c2, then the behaviour of f 12 at c can be analzed through the use of the second derivative test b analzing f 1c2, as follows: a. The graph is concave up, and c is the location of a local minimum value of the function, if f 1c2 7. local minimum f''(c). CHAPTER 4 199

3 b. The graph is concave down, and c is the location of a local maimum value of the function, if f 1c2 6. local maimum f''(c), c. If f 1c2, the nature of the critical point cannot be determined without further work. 4. A point of inflection occurs at 1c, f 1c22 on the graph of f 12 if f 12 changes sign at c. That is, the curve changes from concave down to concave up, or vice versa. (c, f(c)) (c, f(c)) c f'', S d f''. c f''. Sd f'', 5. All points of inflection on the graph of f 12 must occur either where d 2 d equals zero or where 2 is undefined. d 2 d 2 In the following eamples, we will use these properties to sketch graphs of other functions. EXAMPLE 1 Using the first and second derivatives to analze a cubic function Sketch the graph of Solution d d CONCAVITY AND POINTS OF INFLECTION

4 Setting d, d we obtain or 1 d d d Setting 2 we obtain 6 6 or 1. d 2, Now determine the sign of f 12 in the intervals determined b 1. Interval f () 6 7 Graph of f() concave down point of inflection concave up Sketch of f() Appling the second derivative test, at 3, we obtain the local minimum point, 13, 172 and at 1, we obtain the local maimum point, 1 1, 152. The point of inflection occurs at 1 where f The graph can now be sketched. = ( 1, 15) 1 (1, 1) (3, 17) EXAMPLE 2 Using the first and second derivatives to analze a quartic function Sketch the graph of f Solution The first and second derivatives of f 12 are f and f Setting f 12, we obtain 12 2 or CHAPTER 4 21

5 But is also obtained from f 12. Now determine the sign of f 12 on the intervals determined b. Interval 6 7 f () 7 7 Graph of f() concave up? concave up Sketch of f() We conclude that the point 1, 2 is not an inflection point because f 12 does not change sign at. However, since is a critical number and f 12 6 when 6 and f 12 7 when 7, 1, 2 is an absolute minimum. EXAMPLE 3 Using the first and second derivatives to analze a root function Sketch the graph of the function f Solution The derivative of f 12 is f Note that f 12 does not eist, so is a critical number of f 12. It is important to determine the behaviour of f 12 as S. Since f 12 7 for all values of, and the denominator of f 12 is zero when, we have f 12 q. This means that there is a vertical tangent at. In addition, lim S f is increasing for 6 and 7. As a result this graph has no local etrema CONCAVITY AND POINTS OF INFLECTION

6 The second derivative of f 12 is f Since 5 if and , 3 6 if 6, we obtain the following table: Interval 6 7 f () f() does not eist The graph has a point of inflection when, even though f 12 and f 12 do not eist. Note that the curve crosses its tangent at EXAMPLE 4 Reasoning about points of inflection Determine an points of inflection on the graph of Solution The derivative of f is f The second derivative is f f CHAPTER 4 23

7 Setting f 12 gives or ; 1. Determine the sign of f 12 on the intervals determined b 1and 1. Interval f () Graph of f() concave up point of inflection concave down point of inflection concave up Therefore, Q 1, 1 and Q1, 1 are points of inflection on the graph of f R 4 R 1 = 1 ( 1, ( 1, ) 4 1 ) IN SUMMARY Ke Ideas The graph of a function f 12 is concave up on an interval if f 12 is increasing on the interval. The graph of a function f 12 is concave down on an interval if f 12 is decreasing on the interval. A point of inflection is a point on the graph of f 12 where the function changes from concave up to concave down, or vice versa. f 1c2 or is undefined if 1c, f 1c22 is a point of inflection on the graph of f 12. Need to Know Test for concavit: If f 12 is a differentiable function whose second derivative eists on an open interval I, then the graph of f 12 is concave up on I if f 12 7 for all values of in I the graph of f 12 is concave down on I if f 12 6 for all values of in I The second derivative test: Suppose that f 12 is a function for which f 1c2, and the second derivative of f 12 eists on an interval containing c. If f 1c2 7, then f 1c2 is a local minimum value. If f 1c2 6, then f 1c2 is a local maimum value. If f 1c2, then the test fails. Use the first derivative test CONCAVITY AND POINTS OF INFLECTION

8 Eercise 4.4 K PART A 1. For each function, state whether the value of the second derivative is positive or negative at each of points A, B, C, and D. a. A f() b. g() B C D A B C D 2. Determine the critical points for each function, and use the second derivative test to decide if the point is a local maimum, a local minimum, or neither. a c. s t t 1 b. 25 d Determine the points of inflection for each function in question 2. Then conduct a test to determine the change of sign in the second derivative. 4. Determine the value of the second derivative at the value indicated. State whether the curve lies above or below the tangent at this point. w a. f at 2 c. p1w2 at w 3 w 2 1 b. g at 1 d. s1t2 2t at t 2 t 4 PART B 5. Each of the following graphs represents the second derivative, f 12, of a function f 12: a. b. 1 =f''() =f''() f 12 is a linear function. f 12 is a quadratic function. For each of the graphs above, answer the following questions: i. On which intervals is the graph of f 12 concave up? On which intervals is the graph concave down? CHAPTER 4 25

9 C A T ii. List the -coordinates of all the points of inflection. iii. Make a rough sketch of a possible graph of f 12, assuming that f Describe how ou would use the second derivative to determine a local minimum or maimum. 7. In the algorithm for curve sketching in Section 4.3, reword step 4 to include the use of the second derivative to test for local minimum or maimum values. 8. For each of the following functions, i. determine an points of inflection ii. use the results of part i, along with the revised algorithm, to sketch each function. a. b. g1w2 4w2 3 f Sketch the graph of a function with the following properties: f 12 7 when 6 2 and when f 12 6 when 7 5 f 122 and f 152 f 12 6 when 6 2 and when f 12 7 when and when 7 7 f Find constants a, b, and c such that the function f 12 a 3 b 2 c will have a local etremum at 12, 112 and a point of inflection at 11, 52. Sketch the graph of f 12. PART C 11. Find the value of the constant b such that the function f 12 1 b has a point of inflection at Show that the graph of f 12 a 4 b 3 has two points of inflection. Show that the -coordinate of one of these points lies midwa between the -intercepts. 13. a. Use the algorithm for curve sketching to sketch the function b. Eplain wh it is difficult to determine the oblique asmptote using a graphing calculator. 14. Find the inflection points, if an eist, for the graph of f 12 1 c2 n, for n 1, 2, 3, and 4. What conclusion can ou draw about the value of n and the eistence of inflection points on the graph of f? w CONCAVITY AND POINTS OF INFLECTION