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## ← Graphing a Derivative Function | MIT 18.01SC Single Variable Calculus, Fall 2010

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Graphing a Derivative Function

Instructor: Christine Breiner

View the complete course: http://ocw.mit.edu/18-01SCF10

More courses at http://ocw.mit.edu

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Afficher la révision 1 créée 03/19/2011 par audreycermak.

1. Welcome to recitation!
2. Today in this video what we're going to do is
3. look at how we can determine the graph of a derivative of a function from the graph of the function itself.
4. So I've given a function here - we're calling it just y equals f of x.
5. This is the curve y equals f of x.
6. So we're thinking about a function f of x.
7. I'm not giving you the equation for the function.
8. I'm just giving you the graph.
9. And what I'd like you to do, what I'd like us to do in this time,
10. is to figure out what the curve y equals f prime of x will look like.
11. So that's our objective.
12. So what we'll do first is try to figure out the things that we know about f prime of x.
13. So what I want to remind you is that when you think about a function's derivative,
14. remember, its derivative's output is measuring the slope of the tangent line at each point.
15. So that's what we're interested in finding.
16. Understanding the slope of the tangent line of this curve at each x value.
17. So, it's always easiest when you're thinking about a derivative
18. to find the places where the slope of the tangent line is zero
19. because those are the only places where you can
20. hope to change the sign on the derivative.
21. So what we'd like to do is first identify on this curve where
22. the tangent line has slope equal to zero.
23. I think there are two places we can find it fairly easily.
24. That would be at whatever this x value is.
25. That slope there is zero.
26. It's going to be a horizontal tangent line.
27. And at whatever this x value is. The slope there is also zero.
28. Horizontal tangent line.
29. But there's one third place where the slope of the tangent line is zero.
30. It's kind of hidden right in here.
31. Actually, you may think there are a few more, but we're going
32. to assume that this function is always continuing down
33. through this region.
34. So there are three places where the tangent line is horizontal.
35. I can even sort of draw them lightly through here.
36. You have three horizontal tangent lines.
37. So at those points we know that the derivative's value is equal to zero.
38. The output is equal to zero.
39. And now, what we can determine is, between those regions,
40. where are the values of the derivative positive and negative.
41. So what I'm going to do is, below here I'm going to make a line.
42. And we're going to keep track of what the signs of the derivative are.
43. So let me just draw, this will be our sign on f prime.
44. So that's going to tell us what our signs are.
45. So right below, we'll keep track.
46. So here, I'll just come straight down.
47. Here we know the sign of f prime is equal to zero.
48. We know it's equal to zero there.
49. We know it's also equal to zero here.
50. And we know it's also equal to zero here.
51. Now the question is, what is the sign of f prime in this region.
52. (So, to the left of whatever that x value is.)
53. What is the sign of f prime in this region;
54. in this region; and then to the right?
55. So really we can divide up the x values as:
56. left of whatever that x value is;
57. inbetween these two x values;
58. inbetween these two x values;
59. and to the right of this x value.
60. That's really what we need to do to determine what the signs of f prime are.
61. So, again, what we want to do to understand f prime
62. is we look at the slope of the tangent line of the curve y equals f of x.
63. So let's pick a place in this region left of where it's zero.
64. Say, right here. And let's look at the tangent line.
65. The tangent line has what kind of slope? A positive slope.
66. In fact, if you look along here, you see that all of the slopes are positive.
67. So f prime is bigger than zero here.
68. Now I'm just going to record that.
69. I'm going to keep that in mind as a plus, the sign is positive there.
70. Now, if I look right of where f prime equals zero --
71. if I look for x values to the right --
72. I see that as I move to the right, the tangent line is curving down.
73. So let me do it with the chalk.
74. You see the tangent line has a negative slope.
75. If I draw one point in, it looks something like that.
76. So the slope is negative there.
77. So here I can record that the sign of f prime is a minus sign there.
78. Now, if I look between these two x values (which I'm saying
79. here it's zero, and here it's zero for the x values)
80. and I take a point, we notice the sign is negative there also.
81. So, in fact, the sign of f prime changed at this zero of f prime.
82. But it stays the same around this zero of f prime.
83. So, it's negative then it goes to negative again.
84. It's negative, then zero, then negative.
85. And then if I look to the right of this x value and I take a point,
86. I see that the slope of the tangent line is positive.
87. And so the sign there is positive.
88. So we have: the derivative is positive;
89. and then zero; and then negative; and then zero;
90. and then negative; and then zero; and then positive.
91. So there's a lot going on.
92. But if I want to plot now, y equals f prime of x, I have some
93. sort of launching point by which to do that.
94. So what I can do, is I know the that derivative zero --
95. I'm going to draw the derivative in blue here --
96. the derivative is zero (its output is zero) at these places.
97. So I'm going to put those points on.
98. And then if I were just trying to get a rough idea of what happens,
99. the derivative is positive left of this x value.
100. So it's certainly coming down [let me make these a little darker]
101. -- it's coming down because it's positive.
102. It's coming down to zero.
103. It has to stay above the x axis, but it has to head toward zero.
104. Right?
105. What does that actually correspond to?
106. Well, look at what the slopes are doing.
107. The slopes of these tangent lines -- as I move in the x direction --
108. (watch what my hand is doing)
109. the slope is always positive but it's becoming less
110. and less vertical, right? It's headed toward horizontal.
111. The slope that was steeper over here is becoming less steep.
112. The steepness is really the magnitude of the derivative.
113. That's really measuring how far the output is from zero.
114. So as the derivative becomes less steep, the derivative's values
115. have to be headed closer to zero.
116. Now what happens when the derivative is equal to zero here?
117. Well, all of a sudden the slopes are becoming negative.
118. So the outputs of the derivative are negative.
119. It's going down.
120. But then once it hits here again, notice what happens.
121. The derivative is zero again.
122. Notice how I get there.
123. The derivative's negative and the slope of these
124. tangent lines start to get shallower.
125. Right? They were steep, and then somewhere they start to get shallower.
126. So there's some place in the x values between here and here
127. where the derivative is as steep as it gets in this region
128. and then it gets less steep.
129. The steepest point is that point where you have the biggest magnitude
130. in that region for f prime.
131. So that's where it's going to be furthest from zero.
132. So, if I'm guessing, it looks like right around here,
133. the tangent line is as steep as it ever gets in that region,
134. between these two zeros, then it gets less steep.
135. So I'd say right around there, we should say that's as low
136. as it goes, and now it's going to come back up.
137. So hopefully that makes sense.
138. We'll get to see it again here.
139. Between these two zeros, the same kind of thing happens.
140. But notice, we have to be careful.
141. We shouldn't go through zero here because the
142. derivative's output -- the sign is negative, right?
143. So the tangent line -- it was negative, negative, negative, zero
144. oop, it's still negative!
145. So the outputs are still negative, and
146. they're going to be negative all the way to this zero.
147. What we need to see again, is the same kind of thing
148. as happened in this region will happen in this region.
149. The point being that, again, we're zero here -- we're zero here.
150. So somewhere in the middle, we start at zero,
151. the tangent lines start to get steeper.
152. Then at some point, they stop getting steeper.
153. They start getting shallower.
154. That place looks, maybe, right around here.
155. That's the sort of steepest tangent line.
156. Then it gets less steep.
157. So that's the place where the derivative's magnitude
158. is going to be the biggest in this region.
159. Actually, I've sort of drawn it so that they look about the
160. same steepness at those two places.
161. So I should probably put the outputs about the same
162. down here. Their magnitudes are about the same.
163. So this has to bounce off, come up here...
164. (I made that a little sharper than I meant to.)
165. So that's the place -- the tangent line at this x value
166. is the steepest that we get in this region.
167. So the output at that x value is the lowest we get.
168. And then when we're to the right of this zero for the derivative,
169. we start seeing the tangent line's positive,
170. (we pointed that out already)
171. and it gets more positive.
172. So it starts at zero, it starts to get positive,
173. and it gets more positive.
174. So it's going to do something like that, roughly.
175. So let me fill in the dotted line so we can see it clearly.
176. Now this is not exact, but this is a fairly good
177. drawing of f prime of x -- y equals f prime of x.
178. And now I'm going to ask you a question.
179. I'm going to write it on the board.
180. And then I'm going to give you a moment to think about it.
181. So let me write the question.
182. It's, "Find a function g of x so that y equals g prime of x
183. looks like y equals f prime of x."
184. OK, let me be clear about that and then
185. I'll give you a moment to think about it.
186. So I want you to find a function g of x, so that its derivative's graph
187. (y equals g prime of x ) looks exactly like
188. the graph we've drawn in blue here (y equals f prime of x.)
189. Now, I don't want you to find something in terms of
190. x squareds and x cubes.
191. I don't want you to find an actual g of x equals something in terms of x.
192. I want you to just try and find a relationship that it must have with f.
193. So I'm going to give you a moment to think about it
195. And I'll be back to tell you.
196. OK, welcome back!
197. So what we're looking for is a function g of x,
198. so that its derivative, when I graph it,
199. y equals g prime of x, I get exactly the same
200. curve as the blue one.
201. And the point is, if you thought about it for a little bit
202. what you really need is a function that looks
203. exactly like this function y equals f of x,
204. at all the x values in terms of its slopes,
205. but those slopes can happen shifted up or down anywhere.
206. So the point is, if I take the function y equals f of x
207. and I add a constant to it, which shifts the whole graph
208. up or down, the tangent lines are unaffected by that shift.
209. So I get exactly the same picture when I take the
210. derivative of that graph.
211. When I look at the tangent line slopes of that graph.
212. So, you could draw another picture and check it
213. for yourself if you didn't feel convinced.
214. Shift this curve up and then look at what the tangent
215. lines do on that curve.
216. But then you'll see its derivative's outputs are exactly the same.
217. So we'll stop there.