Week #7 Maxima and Minima, Concavity, Applications Section 4.2

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1 Week #7 Maima and Minima, Concavit, Applications Section 4.2 From Calculus, Single Variable b Hughes-Hallett, Gleason, McCallum et. al. Copright 2005 b John Wile & Sons, Inc. This material is used b permission of John Wile & Sons, Inc. SUGGESTED PROBLEMS Find formulas for the functions described in Eercises A cubic polnomial having -intercepts at 1, 5, 7. If we write () in factored form, we can set the desired roots easil: = k( 1)( 5)( 7) 17. Let p() = 3 a, where a is constant and a > 0. (a) Find the local maima and minima of p. (b) What effect does increasing the value of a have on the positions of the maima and minima? (c) On the same aes, sketch and label the graphs of p for three positive values of a. (a) We have p () = 3 2 a. This has values of zero (critical points) at 2 = a/3 or a = ± 3. drawing a sketch of the sign of p (), we have Sign of p () a 3 0 a 3 a From the first derivative test, = 3 local minimum. The actual height of the function at each of these points is a is a local maimum, and = + 3 is a 1

2 a Local ma: p( 3 ) = a a + a a = + 2a a a a a Local min: p( 3 ) = p( 3 ) = 2a 3 3 (b) Increasing the value of a moves the critical points of p awa from the -ais, and moves the critical values awa from the -ais. Thus, the bumps get further apart and higher. At the same time, increasing the value of a spreads the zeros of p further apart (while leaving the one at the origin fied.) Small a Larger a (c) 21. The number, N, of people who have heard a rumor spread b mass media at time, t, is given b N(t) = a(1 e kt ) There are 200,000 people in the population who hear the rumor eventuall. If 10% of them heard it the first da, find a and k, assuming t is measured in das. The epression 200,000 people hear the rumour eventuall can be interpreted as lim t N(t) = 200,000. From the formula for N(t), lim t N(t) = a(1 0) = a so a = 200,000. If after one da, the fraction that heard the rumour (at t = 0) is 10% of 200,000, N(1) = ,000 = 200,000(1 e k(1) ) 0.1 = 1 e k e k = 0.9 k = ln(0.9)

3 26. (a) Find all critical points of f() = 4 + a 2 + b. (b) Under what conditions on a and b does this function have eactl one critical point? What is the one critical point, and is it a local maimum, a local minimum, or neither? (c) Under what conditions on a and b does this function have eactl three critical points? What are the? Which are local maima and which are local minima? (d) Is it ever possible for this function to have two critical points? No critical points? More than three critical points? Give an eplanation in each case. (a) f () = a = 2(2 2 + a) Setting f () = 0 to find the critical points, = 0 and = ± a/2 are critical points. (b) = 0 is alwas a critical point for f(). However, the other values will onl be critical points if a < 0. If a < 0 then we will have three real critical points, otherwise (for a 0), we will onl have the one real critical point at = 0. The value of b is just a vertical shift on the graph of f(), and has no effect on the number of critical points. The sign of f () will be negative to the left of = 0, and positive to the right, so = 0 is a local minimum. (c) As indicated in part (b), if a < 0 then we will have three real critical points, otherwise (for a 0), we will onl have the one real critical point at = 0. A sketch of the sign of f () is shown below. Sign of f ()=2(2^2 + a) a 2 0 a 2 From the first derivative test, = a/2 and = a/2 are local minima, while = 0 is a local maimum. (d) For a 0, there is eactl one critical point, = 0. For a < 0 there are eactl three different critical points. These ehaust all the possibilities. (Notice that the value of b is still irrelevant.) 43. An organism has size W at time t. For positive constants A, b, and c, the Gompertz growth function gives W = Ae eb ct,t 0 3

4 (a) Find the intercepts and asmptotes. (b) Find the critical points and inflection points. (c) Graph W for various values of A, b, and c. (d) A certain organism grows fastest when it is about 1/3 of its final size. Would the Gompertz growth function be useful in modeling its growth? Eplain. (a) The vertical intercept is W = Ae eb c 0 = Ae eb. There is no horizontal intercept since the eponential function is alwas positive. There is a horizontal asmptote. As t, we see that e b ct = e b /e ct 0, since t is positive. Therefore W Ae 0 = A, so there is a horizontal asmptote at W = A. (b) The derivative is dw dt = ( Ae eb ct) ( e b ct) ( c) = cae eb ct e b ct Thus, dw/dt is alwas positive, so W is alwas increasing and has no critical points. The second derivative is d 2 W dt 2 = d ( cae eb ct) e b ct + cae d eb ct dt dt eb ct ( ) = cae eb ct ( e b ct )( c) e b ct + cae eb ct (e b ct )( c) ( = c 2 Ae eb ct e b ct)( ) e b ct 1 Now e b ct decreases toward 0 as t. The second derivative changes sign from positive to negative when e b ct = 1, i.e., when b ct = 0, or t = b/c. Thus the curve has an inflection point at t = b/c, where W = Ae eb (b/c)c = Ae 1. A=50, b=2, c=1 A=50, b=2, c=1 W A=20, b=2, c=1 (c) time 4

5 (d) The final size of the organism is given b the horizontal asmptote, W = A. The curve is steepest at its inflection point, which occurs at t = b/c, W = A e 1. Since e = , the size the organism when it is growing fastest is about A/3, one third of its final size. So es, the Gompertz growth function is useful in modeling such growth. QUIZ PREPARATION PROBLEMS Find formulas for the functions described in Eercises A function of the form = Asin(B)+C with a maimum at (5, 2), a minimum at (15, 1.5), and no critical points between these two points. Ma-min range of 1.5 to 2 means the amplitude, A, is (2-1.5)/2 = Center of the oscillations is 1.75, so C = Time from min to ma is 15-5=10, which is half a period. The full period must be 20, so B = 2π/20 = π/10. So = 0.25sin( π ) A curve of the form = e ( a)2 /b for b > 0 with a local maimum at = 2 and points of inflection at = 1 and = 3. This is a transformed version of e 2 which is the bell-shaped curve we have seen previousl. The a shifts it to the right b a, and dividing b b produces a horizontal contraction or stretch. e 2 has its local ma at = 0, so = e ( a)2 /b has its local ma at = a (shifted right b a). Since we are told the maimum is at = 2, we must have a = 2. Tracking down the points of inflection requires differentiating. d 2) = 2( e ( 2)2 /b d b d 2 ( d 2 = 2 ( e b) ) ( ) ( 2)2 /b 2( 2) ( )( ) e ( 2)2 /b 2( 2) b b ( 2 /b = )( 1 b e ( 2)2 + 2b ) ( 2)2 Since e ( 2)2 /b is never zero, d 2 /d 2 = 0 where b ( 2)2 = 0 We know that d 2 /d 2 = 0 at = 1, so substituting = 1 gives 5

6 1 + 2 b ( 1)2 = 0 Solving for b gives 2 b = 1 b = 2 With a = 2 and b = 2, our function is = e ( 2)2 /2 35. Sketch several members of the famil e a sin(b) for a = 1, and describe the graphical significance of the parameter b. If were Asin(b), we would sa that the were oscillations with different periods, but all with amplitude (vertical up + down range) of A. Having an eponential instead of the A simpl means that the amplitude changes with time. Think of the oscillations of a spring or a pendulum, and how the will graduall get smaller and smaller over time. The larger the value of b, the faster the oscillations. 36. Consider the famil = A + B (a) If B = 0, what is the effect of varing A on the graph? (b) If A = 1, what is the effect of varing B? 6

7 (c) On one set of aes, graph the function for several values of A and B. (a) The larger the value of A, the steeper the graph (for the same -value). (b) The graph is shifted horizontall b B. The shift is to the left for positive B, to the right for negative B. There is a vertical asmptote at = B. (c) We ll use several aes for clarit. See the graphs below. A=1,B=0 A=5,B=0 A=1,B=2 A=5,B=2 7

Week #6 - Taylor Series, Derivatives and Graphs Section 4.1

Week #6 - Taylor Series, Derivatives and Graphs Section 4.1 Week #6 - Talor Series, Derivatives and Graphs Section 4.1 From Calculus, Single Variable b Hughes-Hallett, Gleason, McCallum et. al. Copright 2005 b John Wile & Sons, Inc. This material is used b permission