Part 4 Review Part 2 — Transcript

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1. 0:00Hello, this is chapter 4 review part
2. 0:02two. So number six says find the second
3. 0:05derivative of y =x^2 + x. To take the
4. 0:09second derivative, we take the
5. 0:10derivative once and then we take the
6. 0:12derivative again. So the first
7. 0:14derivative would be y prime. The
8. 0:16derivative of -x^2 would be -2x.
9. 0:20Derivative of 1x would be 1. So the
10. 0:22first derivative would be -2x + 1. To
11. 0:25find the second derivative yblep prime,
12. 0:28the derivative of -2x is -2 and the
13. 0:30dative of 1 is 0. So the second
14. 0:32derivative yblep prime would equal -2.
15. 0:36Question seven, find the second
16. 0:39derivative. So we have y = -2x^2
17. 0:43+ 3x 6 - 4x. I would step one rewrite
18. 0:47it. So instead of having -2x^2,
19. 0:50I would do y = -2x -2 + 3x 6 - 4x. Then
20. 0:57we take the first derivative y prime.
21. 0:59The derivative of -2x -2 would be 4x -3.
22. 1:04The derivative of 3x 6 would be 18 x
23. 1:075th. And the dative of - 4x would be -4.
24. 1:11Then we take the second derivative which
25. 1:13would be the derivative of 4x^ -3 would
26. 1:16be -12x4.
27. 1:19The derivative of 18x 5th would be uh
28. 1:2290x 4. And the dative of -4 is 0. So our
29. 1:27second derivative would be -12. You can
30. 1:30take that x^4 move it to the
31. 1:32denominator. So it' be -12 x 4 + 90x 4.
32. 1:39Okay. Question eight, find the second
33. 1:42derivative of y = ex + e x. So the first
34. 1:46derivative, the derivative of e^x,
35. 1:49you step one, rewrite it, egx. And then
36. 1:52the d of the exponent, the d of -1x is
37. 1:55-1 plus the d of e x is e x. So we'll
38. 2:00simplify along the way. So instead of
39. 2:02having ex *1, we can say the first
40. 2:04derivative is e x + e x. Then take the
41. 2:09derivative. So the negative stays in
42. 2:11front. Rewrite egx egx. And the dative
43. 2:15of -1x is -1 plus the d of e x e x. Then
44. 2:20we'll simplify again.
45. 2:221 would be positive ex + e x.
46. 2:28Question 9. The following formula gives
47. 2:31the oxygen consumption in milliliters of
48. 2:33a bird embryo for the time the egg is
49. 2:35laid through the bird hatches in days.
50. 2:37So we have O of T =0.065
51. 2:42T cub + 3.4 T ^2 - 22T + 3.6 from T of 8
52. 2:49to 30. So remember our input T is in
53. 2:51days and our output is oxygen
54. 2:53consumption in millilit. So uh we're
55. 2:56given that the window for X is 8 to 30
56. 2:59days 8 to 30 and then the minimum is
57. 3:02from 0 to a,000. So our so our y so our
58. 3:05y minimum is zero our y maximum is a
59. 3:08th00and and then we graph the original
60. 3:10function you can see how it changes.
61. 3:12So if I want to find the oxygen
62. 3:14consumption of day 20 you take 20 plug
63. 3:17into the original function and you get
64. 3:19403.6.
65. 3:21So at day 20 the oxygen consumption is
66. 3:23403.6 millilit. If I'd like to find the
67. 3:26rate of change of the oxygen consumption
68. 3:28I would take the derivative the first
69. 3:31derivative of the original. So the
70. 3:34original is0.065
71. 3:36t cub + 3.4 t ^2 - 22t + 3.6. The
72. 3:41derivative of that
73. 3:44would be0.195
74. 3:46t ^2 + 6.8 tus 22. And what is the rate
75. 3:51of change of the oxygen consumption on
76. 3:52day 20? We plug 20 into the first
77. 3:54derivative and we get 36. So at day 20
78. 3:58the rate of change of the oxygen
79. 4:00consumption is increasing 36 milll per
80. 4:02day. Now what is the function to find
81. 4:05the rate of oxygen consumption's
82. 4:07acceleration? So when I say acceleration
83. 4:10that's second derivative. So we take the
84. 4:12derivative again. So the derivative
85. 4:14of0.195t
86. 4:16^2 + 6.8 tus 22 would be0.39t
87. 4:22+ 6.8.
88. 4:25And then when we plug 20 into the second
89. 4:28derivative we get neg -1. So at day 20
90. 4:31the oxygen consumption is decelerating
91. 4:34at 1 milll per day squared per day
92. 4:37squared.
93. 4:40Okay. Question 10. Given the function
94. 4:43f(x) = 3x 3r + 9x^2 - 720x - 15 from -12
95. 4:50to 10 would be the domain. Find the x
96. 4:52intercepts and y intercepts. So remember
97. 4:53to find the e any x intercepts you make
98. 4:56y zero. So we take the original function
99. 5:00set y equal to zero and I could use the
100. 5:02calculator for this. If you plug the
101. 5:04original in y1 3x cub + 9x^2 - 720x - 15
102. 5:10and y 2 set zero in an appropriate
103. 5:13window which I'll show below and you
104. 5:15find that the intersection is0.0210.
105. 5:19That would be your x intercept.
106. 5:22To find the y intercept you make x zero.
107. 5:24So if I plug 0 into the original
108. 5:26remember x intercept y intercept go back
109. 5:28to the original you get -15. So we have
110. 5:30a y intercept at 0 comma -15.
111. 5:34Then it asks are there any vertical
112. 5:36asmtotes or horizontal asmmptotes.
113. 5:38Remember for vertical asmmptotes we see
114. 5:40if there's any place where it's the
115. 5:41original is undefined. None.
116. 5:45And then for horizontal asmmptotes again
117. 5:47uh we have those three possibilities for
118. 5:49horizontal asmmptotes. you may want to
119. 5:51review. But in this case, we're not
120. 5:52dividing. There's no numerator,
121. 5:53denominator or two functions. So there's
122. 5:55no horizontal asmmptotes. This is a
123. 5:57continuous function.
124. 6:00To do stationary points, we take the
125. 6:02derivative, set it equal to zero. So the
126. 6:03derivative is 9x^2 + 18x - 720= z. So we
127. 6:08take the derivative of the original, set
128. 6:10it equal to zero. Uh you can factor out
129. 6:13the 9. And then you have x + 10 * x - 8.
130. 6:16So we have two answers. x= -10, x= 8 or
131. 6:21we could use quadratic formula where a
132. 6:23is 9, b is 18, c is -7,20. So possible
133. 6:26stationary points are -10, 8.
134. 6:30And then singular, take the derivative,
135. 6:32see if there's any place where it's
136. 6:33undefined. None. No number would make it
137. 6:36undefined. No singular points and end
138. 6:38points given by the domain -12 and 10.
139. 6:43And if we graph the original function,
140. 6:46which is 3x 3r + 9x^2 - 720x - 15 from
141. 6:50-12 to 10, looking at the y values that
142. 6:54we're looking at. So when you plug in
143. 6:57-10 into the original, you get 5,85. 8
144. 6:59into the original, -3,63.
145. 7:03-12 into the original, you get 4,737.
146. 7:07Plug 10 into the original, you get
147. 7:08-3,315.
148. 7:11Looking at the y values, uh, the lowest
149. 7:14is -3,663,
150. 7:16so I went to -4,000. The highest is
151. 7:195,850,
152. 7:21so I went to 6,000. And then we'll look
153. 7:23at each point. So -10, 5,85.
154. 7:28That would be an absolute max.
155. 7:32The next point 8 comma uh 3,663
156. 7:39that would be an absolute minimum.
157. 7:43The next point uh -124737
158. 7:50that would be a minimum but it's not the
159. 7:52lowest point in the entire graph. So
160. 7:54that would be a local minimum.
161. 7:57And then 10 comma -315
162. 8:01which would be here would be a maximum
163. 8:06but is not the highest point of the
164. 8:07paragraph. So that would be a local max.
165. 8:10To find points of inflection you take
166. 8:12the second derivative. So remember the
167. 8:14first derivative is 9x^2 + 18x - 720.
168. 8:18The second derivative of the derivative
169. 8:20of 9x^2 is 18x. Derative of 18x is 18.
170. 8:24Derative of -720 is zero. Then we set
171. 8:2718x + 18 the second derivative equal to
172. 8:30zero.
173. 8:32Subtract 18. Subtract 18 to both sides.
174. 8:34Negative sorry 18x =8.
175. 8:37Divide both sides by 18. x=1.
176. 8:41We take1 plug it into the original
177. 8:43function. We get 712. So we have an
178. 8:45inflection point at -1a 712. 7-11
179. 8:52comma 711 would be here. That's where it
180. 8:54changes from concave down to concave up.
181. 8:57So we can verify that with a graph. So
182. 8:59we have an inflection point at -1, 711.
183. 9:03When is the graph concave up? If you
184. 9:05look at the graph, it is concave up from
185. 9:07that -1 on to 10. And when is it concave
186. 9:12down? It's concave down from -12 to -1.
187. 9:18Notice it includes -12 parenthesy of -1.
188. 9:21That's why it's concave down. and then
189. 9:23parenthesy at -1 all the way to 10
190. 9:26including -10 bracket is when it's
191. 9:28concave up.
192. 9:31Okay, question 11. Given the function uh
193. 9:34x^2 + 1/x^2, we'll go through the same
194. 9:37steps again this one. So to find the x
195. 9:39intercepts, you make y 0. So if we set
196. 9:41the original equal to zero, x^2 + 1x^2 =
197. 9:440. To solve algebra, we subtract 1/x^2
198. 9:48subtract 1x^2.
199. 9:50So we have x^2 =1x^2 cross multiply x^2
200. 9:54* x^2 is x 4 1 * 1 is1 we have x 4 =1 x^
201. 10:00the 4th power cannot equal ne1 so
202. 10:02there's no solution therefore no x
203. 10:03intercept to find a y intercept you make
204. 10:06x0 if I plug 0 into the original uh you
205. 10:10get undefined so therefore there's no y
206. 10:12intercept
207. 10:13looking at the graph again we can see if
208. 10:15there's any vertical asmtotes looking at
209. 10:17the original
210. 10:18the zero would make it undefined So we
211. 10:20have a vertical asmtote at zero.
212. 10:23And then for horizontal asmtote uh we
213. 10:26could simplify the original to have x 4
214. 10:28+ 1 /x^2. Basically trying to find a
215. 10:32common denominator here. This would be x
216. 10:344 over x^2 + 1x^2. So x 4 + 1 over x^2.
217. 10:43And remember your rules for horizontal
218. 10:45asmmptotes. If the degree of the
219. 10:46numerator is larger than the degree of
220. 10:48the denominator, there's no horizontal
221. 10:50asmtote.
222. 10:52Then for the extrema, we take the first
223. 10:55derivative of the original. Uh the
224. 10:57original is x^2 + x^ -2, the derivative
225. 11:00would be 2x - 2x -3. Set that equal to
226. 11:03zero. So if you had 2x - 2x 3r = 0. Add
227. 11:082x 3r to both sides. 2x = 2x cubed.
228. 11:13Cross multiply since we have a
229. 11:14proportion. 2x * x cub is 2 x 4. 2 * 1
230. 11:18is 2. 2 x 4 = 2. Divide both sides by 2.
231. 11:22X 4 = 1. Therefore, x= plus or minus 1.
232. 11:26So you have two possible 1 and 1.
233. 11:30Singular points. What would make the
234. 11:31derivative undefined? 0. And end points,
235. 11:34none were given. So we have 01 and one.
236. 11:39Remember we plug those into the
237. 11:40original. So negative 1 into the
238. 11:42original would give us two.
239. 11:44one into the original would give us two
240. 11:46and zero into the is undefined. So it
241. 11:48wouldn't be a max or min. If we graph
242. 11:50the original from 5 to 5 and from 0 to
243. 11:54negative sorry 0 to 5 this would be our
244. 11:57graph. Notice the two points that we
245. 12:00have -12 and 1 2 are both minimums.
246. 12:04They're both the lowest. So that would
247. 12:06be we have both of them being absolute
248. 12:09minimums. Absolute minimums. To find
249. 12:12point of inflection, you take the second
250. 12:14derivative. So we have our original x^2
251. 12:17+ x -2. The first derivative is 2x - 2x
252. 12:21-3. The second derivative would be
253. 12:24derivative of 2x is 2. Div -2x -3 is +
254. 12:286x4.
255. 12:30Set that equal to 0. 2 + 6x 4 = 0.
256. 12:34Subtract 6x^2. Subtract 6x^2. 2 = -6x 4.
257. 12:39Sorry. um subtract 6 to x 6 / x 4 - 6 x
258. 12:454 which gives us 2 = -6 / x 4 cross
259. 12:49multiply 2 * x^ 2 * x 4 = 1 * -6 so 2 x
260. 12:564 = -6 divide both sides by 2 x 4 = -3 x
261. 13:024 cannot equal -3 so no solution no
262. 13:05points in if you look at the graph it
263. 13:07does not go from concave up to concave
264. 13:10down. It's concave up, concave up. So
265. 13:13when is the graph concave up? From
266. 13:15negative infinity to 0 and from 0 to
267. 13:20infinity.
268. 13:22And when is a concave uh graph down,
269. 13:24it's not concave down. Thank you.