6.3. Concavity and the Second Derivative Test

Transcription

1 6.3 Concavit and the Second Derivative Test Imagine driving along a curved road from R to T as shown. As the car moves from R to S, the road curves to the left, and ou feel the seat of the car eert an additional force on ou to the left. This force tends to decrease as ou approach S, and reaches zero at S. The road then curves to the right and the additional force from the seat points to the right. In this section, we eplore curves that behave in this fashion, and points such as S. T S T S R R If we think of the road as the graph of a function, then the curve from R to S is described as concave upward (curving so as to enclose the space above it). From S to T the curve is concave downward, and from T on it is concave upward. To distinguish between these two tpes of curves, eamine the graphs of the functions f and g below. Each graph connects point A to point B, but the two graphs bend in different was. If we draw tangents to the curves at various points, the graph of f lies above its tangents and is called concave upward, while the graph of g lies below its tangents and is called concave downward. B A = f () = g () A a b Concave upward on (a, b) B a b Concave downward on (a, b) The graph of f is said to be concave upward on an interval (a, b) if it lies above all of its tangents on (a, b). The graph of f is said to be concave downward on (a, b) if it lies below all of its tangents on (a, b). 33 MHR Chapter 6

2 For instance, the function shown below is concave upward on the intervals (, ), (3, 5), and (5, 6.5), and concave downward on (, 3) and (6.5, ). P Q R S A point P on a curve is said to be a point of inflection if the curve changes from concave upward to concave downward or from concave downward to concave upward (that is, it changes concavit) at P. For instance, the graph just considered has three points of inflection, namel, P, Q, and S. Point R is a cusp but is not a point of inflection, since the concavit does not change at R. Also, notice that if a curve has a tangent at a point of inflection, as at points P, Q, and S in the figure, the tangent crosses the curve there. In the following investigation, ou will eplore the connection between concavit and tangents. Investigate & Inquire: Concavit and Inflection Points. Graph each function. Estimate the intervals on which the function is concave upward, the intervals on which it is concave downward, and the point(s) of inflection. a) f () 3 3 b) f () c) f () d) f () Calculate f () and f () for each function in step. Graph the function and its derivatives on the same set of aes. 3. For each graph in step, determine the sign of f () on each interval on which the graph is concave upward. Repeat this for each interval on which the graph is concave downward.. Make a conjecture about the sign of f () when the graph is concave upward. Repeat when the graph is concave downward. 5. Draw a vertical line through the point of inflection of each graph of f in step. Let the -coordinate of the point of inflection be a. 6. How does the first derivative behave near an inflection point (a, f (a))? Conjecture a test for inflection points using onl the first derivative. Test our conjecture. 7. How does the second derivative behave near an inflection point (a, f (a))? Conjecture a test for inflection points using onl the second derivative. Test our conjecture. 8. What is the value of the second derivative at each inflection point ou found? 6.3 Concavit and the Second Derivative Test MHR 33

3 Note from the investigation that the sign of the second derivative affects the direction of concavit. If the second derivative is positive, the graph is concave upward. If the second derivative is negative, the graph is concave downward. This can be understood b thinking of the second derivative as the rate of change of the slope of a graph. Thus, if f (), the slope of the graph of f is increasing at. Eamine the following graph of a function that is concave upward on (a, b). We can see that the slopes of the tangents to the function are increasing as increases. The slopes on the = f( ) left are negative, increasing through negative values up to, and then increasing through positive values. A similar argument can be made for a function that is concave downward on an interval. Test for concavit: If f () for all (a, b), then the graph of f is concave upward on (a, b). If f () for all (a, b), then the graph of f is concave downward on (a, b). 3 5 f ( ) f ( ) f ( 3 ) f ( ) f ( 5 ) It follows from the test for concavit and the definition of a point of inflection, that there is a point of inflection at an point where the second derivative changes sign. Note that the second derivative must be zero or not eist at a point of inflection. This compares closel with Fermat s theorem (page 3), which relates critical numbers to the first derivative. As with Fermat s theorem, the condition f or does not eist does not guarantee that a point of inflection eists, as Eample shows. Eample A Function With Zero Second Derivative but no Point of Inflection Show that the function f () satisfies f () but has no point of inflection. Solution Since f (), we have f () 3 and f (), so f (). But for both and, so the concavit does not change and there is no point of inflection. 6 8 f( ) = 3 3 Eample Concavit and Points of Inflection for a Cubic Function a) Determine where the curve f () is concave upward and where it is concave downward. b) Find the points of inflection. c) Use this information to sketch the curve. Draw the tangent at the point of inflection. 33 MHR Chapter 6

4 Solution a) If f () then f () and f () 6 6 6( ) We use an interval chart to determine when f () is negative and when it is positive. The curve is concave downward (since f () ) for (, ) and concave upward (since f () ) for (, ). b) The curve changes from concave downward to concave upward when, so the point (, 5) is a point of inflection. c) Net, we determine the critical numbers of f (). Since f is continuous, we need onl consider -values for which f () = ± ( 6) ( 6) ( 3)( 5) 3 () 6± = 6 Intervals Test values Signs of f( ) Nature of graph concave downward f ( ) = 6( ) (, ) (, ) + concave upward Since there is no real value of for which f (), the function has no critical points. Hence, the function is either alwas increasing or alwas decreasing. We test a particular point to determine which. Since f () 5, the function is alwas increasing. This information, together with parts a) and b) and the -intercept, f (), allows us to sketch the curve. We also need to determine the equation of the tangent at the point of inflection. The slope of the tangent is m f() 3 () 6 () 5 The equation of the tangent is m( ) 5 ( ) f( )= (, 5) = + 3 Note that the information regarding concavit is ver helpful in sketching the curve. 6.3 Concavit and the Second Derivative Test MHR 333

5 Eample 3 Concavit and Points of Inflection for a Rational Function Discuss the curve = with respect to concavit and points of inflection. + Solution Let f( )=. + Then f( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) f ( ) ( ) 3 ( ) 3 ( ) We determine when f is negative and when it is positive in order to determine the concavit. To do this, we first determine when f (). 3 ( ) = 3 ( + ) We can multipl both sides of the equation b the denominator, which is alwas positive. (3 ) 3 =± 3 Intervals Test values Signs of f( ) Nature of graph ( 3 concave upward f( ) = ( + ) (, 3 3 ( 3 ( (3 ) + + concave downward ( concave upward f ( )= + (.58,.75) (.58,.75) ( We consider the intervals,, and. 3,,, The curve is concave upward for and, and concave, 3 3, downward for. The points of inflection are, and,., The graph verifies these calculations. 33 MHR Chapter 6

6 The Second Derivative Test The following investigation eplores the possibilit of using the second derivative of a function to determine if a local etremum is a local maimum or a local minimum. Investigate & Inquire: Distinguishing Local Maima from Local Minima. Eamine the graphs of the functions f, f, and f from step in the investigation on page 33. Draw vertical lines through the local minimum and maimum points of each graph of f.. How do the derivatives f and f behave at these local etrema? 3. Can the second derivative be used to distinguish local maimum points from local minimum points? If so, conjecture a general rule.. Test our rule on two additional functions. Note, from the investigation, that the second derivative can be used to determine whether a local etremum is a maimum or a minimum value of a function f. We assume that f () eists and is continuous throughout the domain of f. The figure below shows the graph of a function f with f (c) and f (c). Since f (c), the graph of f is concave upward near c, and therefore lies above its tangents at (c, f (c)). But since f (c), this tangent is horizontal. Therefore, f has a local minimum at c. = f( ) (, c f()) c c f( c) =, f( c) > Similarl, as shown in the figure on the right, if f (c) and f (c), then the graph of f is concave downward near c, and therefore lies below its horizontal tangent at (c, f (c)). Thus, f has a local maimum at c. = f( ) (, c f()) c c f( c) =, f( c) < 6.3 Concavit and the Second Derivative Test MHR 335

7 Second derivative test: If f (c) and f (c), then f has a local minimum at c. If f (c) and f (c), then f has a local maimum at c. Eample Local Maimum and Minimum Values, Concavit, and Points of Inflection 3 Find the local maimum and minimum values of the function =. Use these, together with concavit and points of inflection, to sketch the curve. Solution 3 Let f( )=. Then f () 3 6 ( 6) f () 3 3( ) To find the critical numbers, we set f () and obtain and 6. Then, to use the second derivative test, we evaluate f at these numbers: f () and f (6) 36 Since f (6) and f (6), f (6) 8 is a local minimum. Since f (), the second derivative test gives no information about the critical number. But, since the first derivative does not change sign at (it is negative on both sides of ), the first derivative test tells us that f has no maimum or minimum at. Since f () 3( ), f () at and at. We determine the concavit using an interval chart as shown above to the right. Intervals Test values Signs of f( ) Nature of graph concave upward f ( )=3 ( ) (, ) (, ) (, ) concave downward concave upward point of inflection (, ) ) f ( )= ) point of inflection (, 6) Thus, f is concave upward on (, ) and (, ) and concave downward on (, ). The points of inflection are (, ) and (, 6). Now, we graph the function using this information. 8 local minimum (6, 8) 336 MHR Chapter 6

8 Eample shows that the second derivative test gives no information when f (c). It also fails when f (c) does not eist. In such cases, we must use the first derivative test. Ke Concepts The graph of f is called concave upward on an interval (a, b) if it lies above all of its tangents on (a, b). It is called concave downward on (a, b) if it lies below all of its tangents on (a, b). A point P on a curve is called a point of inflection if the curve changes concavit at P. Test for concavit If f () for all on (a, b), then the graph of f is concave upward on (a, b). If f () for all on (a, b), then the graph of f is concave downward on (a, b). At a point of inflection, f or f does not eist. However, this condition on f ma also be true at other points. Therefore, inflection points cannot be located simpl b setting f (). A point of inflection for f occurs when the sign of f changes at that point. Second derivative test If f (c) and f (c), then f has a local minimum at c. If f (c) and f (c), then f has a local maimum at c. The second derivative test does not appl when f (c) and when f (c) does not eist. Communicate Your Understanding. Referring onl to the characteristics of a graph, describe the terms a) concave upward b) concave downward. Draw two different curves, each with a point of inflection, so that one of the curves is differentiable at its point of inflection and the other is not. Mark each point of inflection and eplain wh it is a point of inflection. 3. Eplain each statement. a) If f is concave upward on an interval, then the slopes of the tangents are increasing from left to right on that interval. b) If f is concave downward on an interval, then the slopes of the tangents are decreasing from left to right on that interval.. Chris claims that, to find inflection points, all ou have to do is find where the second derivative is or undefined. Discuss the validit of this statement. 5. If f (c) and f (c), what can ou conclude about the point on the graph of f where c? 3 6. Does the second derivative test appl to the function f( )= at? If es, appl it. If no, eplain how to determine whether f has a maimum value, a minimum value, or neither at. 6.3 Concavit and the Second Derivative Test MHR 337

9 A Practise. a) State the intervals on which f is concave upward and the intervals on which it is concave downward. b) State the coordinates of the points of inflection. = f( ) d) State the -coordinates of an points of inflection of f. Eplain wh each is a point of inflection. = f ( ) The figure is a sketch of f for a function f. State the intervals on which the graph of f is concave upward and the intervals on which it is concave downward. Determine the -coordinates of an inflection points of the graph of f. = f( ) 6 8. The following are graphs of the first and second derivatives of a function f (). For each graph, i) identif all the critical numbers of f and eplain if the relate to a maimum, a minimum, or ou cannot decide, using onl the second derivative test ii) for the numbers for which the second derivative test fails, use the first derivative test to decide the nature of the related point a) = f ( ) 6 = f ( ) 3. The graph of the first derivative f of a function f is shown. a) On which intervals is f increasing? decreasing? b) On which intervals is f concave upward? concave downward? c) State the -coordinates of an local maimum or minimum points of the graph of f MHR Chapter 6

10 b) c) = f ( ) = f ( ) 6 8 = f ( ) = f ( ) 5. Find the intervals on which each curve is concave upward and the intervals on which it is concave downward. State an points of inflection. a) f () 3 5 b) c) g() d) 3 3 e) f () f) g) h() h) i) f( )= 3 j) k) = g ( )= + 6. Use the second derivative test to find the local maimum and minimum values of each function, wherever possible. B a) 5 8 b) f () 3 c) g() 3 d) f () 3 9 e) 3 f) f () g) h() ( 8 ) h) 7. Find the local maimum and minimum values of each function. a) g() 6 b) = ( ) c) h ( )= d) Use the first and second derivative tests to determine the maimum points, minimum points, and points of inflection for each function. Graph each function. a) f () 6 b) f () 3 6 c) 3 6 d) f () e) 6 3 f) g) h) Appl, Solve, Communicate 9. a) Determine where f is positive, where it is negative, and where it is zero. Repeat for f and f. b) Is there an interval where f, f, and f? If so, state the interval = f( ) 3 3 h ( ) 6.3 Concavit and the Second Derivative Test MHR 339

11 . A manufacturer keeps accurate records of the cost, C(), of producing items in its factor. The graph shows the cost function. = C ( ) a) Eplain wh C(). b) Eplain the significance of the point of inflection. c) Use the graph of C to sketch the graph of the marginal cost function, C ().. Application A tpical predator-pre relationship for foes and rabbits is shown in the graph.. Communication A software compan estimates that it will sell N units of a new product after spending dollars on TV ads during football games. The relationship can be modelled b N(), [, 5], where is measured in thousands of dollars. a) Find the minimum and maimum number of sales. b) Does more advertising alwas lead to more sales? Eplain. 3. Inquir/Problem Solving Water is poured into a cone-shaped vase, as shown, at a constant rate. a) Sketch a rough graph of H(t), the height of the water in the vase as a function of time. b) Sketch rough graphs of H(t) and H(t). c) Eplain the shape of all three graphs in terms of the shape of the vase. Eplain the concavit in the graph of H(t). Number of rabbits 8 (, 88) Ht () 6 Number of foes 8 6 Time (das) 3 3. For the function f( ) = ( + ), determine a) the intervals of increase and decrease b) the local maimum and minimum values c) the intervals of concavit d) the points of inflection a) Estimate the intervals on which the graph of the pre population is concave upward; concave downward. b) Estimate the coordinates of the points of inflection of the pre graph. c) Describe the predator-pre relationship in terms of maimum and minimum values, intervals of increase and decrease, and concavit. 5. Inquir/Problem Solving If possible, sketch the graph of a continuous function with domain R satisfing each set of characteristics. If it is not possible, eplain wh not. a) The first and second derivatives are alwas positive. b) The function is alwas positive and the first and second derivatives are alwas negative. 3 MHR Chapter 6

12 C c) The first derivative is alwas positive and the second derivative is alwas negative. d) The first derivative is alwas negative and the second derivative is alwas positive. e) The first derivative is alwas positive and the second derivative alternates between positive and negative. f) The function is alwas negative and the first and second derivatives are alwas negative. 6. Application Use a graphing calculator or graphing software. Give results to three decimal places. i) Draw the graph of f. ii) Find all etrema for f. iii) Find all points of inflection for f. a) f () b) 3 ( + ) f( ) = ( ) ( ) 7. Is it possible to use the graph of the second derivative of a function to determine all the -coordinates of the maimum and minimum points without an knowledge of the first derivative? Eplain. 8. For what values of the constants c and d is (, 3) a point of inflection of the cubic curve 3 c d? 9. Graph several members of the famil of polnomials defined b f () c. a) For which values of c do the curves have maimum points? b) Prove that the minimum and maimum points of each of these curves lie on the parabola defined b.. Sketch the graph of a continuous function that satisfies all these conditions. f () f (3), f () f() f () f () f () for (, ) and for (, ) f () for (, ) and for (, ) f () for 3 ( ), f () for (3, ) lim f( ), lim f( ). Sketch the graph of a continuous function that satisfies all of the following conditions. f () for (, ), f () for (, ) f () for (, ), f () for (, ) lim f( ) f is an odd function. Communication The function f () ( )( ) was graphed with technolog. a) Use calculus techniques to determine an local etrema and points of inflection for this curve. b) Do the window settings chosen show all the relevant features of the graph? Eplain. c) If our answer in part b) was no, give appropriate window settings for graphing this function using technolog. 6.3 Concavit and the Second Derivative Test MHR 3