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Title: Intermediate Value Theorem
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Series: Real Analysis
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Chapter: Continuous Functions
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YouTube-Title: Real Analysis 32 | Intermediate Value Theorem
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Bright video: https://youtu.be/BNLu4_3Okuk
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Quiz: Test your knowledge
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Exercise Download PDF sheets
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Thumbnail (bright): Download PNG
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Subtitle on GitHub: ra32_sub_eng.srt
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Timestamps
00:00 Intro
00:14 Introducing the Intermediate Value Theorem
01:26 Definition Intermediate Value Theorem
02:29 Corollary
02:56 Proof of the Intermediate Value Theorem
08:16 Credits
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Subtitle in English
1 00:00:00,349 –> 00:00:02,329 Hello and welcome to real
2 00:00:02,339 –> 00:00:03,349 analysis.
3 00:00:03,720 –> 00:00:05,260 And as always, first, I really
4 00:00:05,269 –> 00:00:06,449 want to thank all the nice
5 00:00:06,460 –> 00:00:07,409 people that support this
6 00:00:07,420 –> 00:00:08,939 channel on Steady or pay pal.
7 00:00:09,609 –> 00:00:10,970 And now in today’s part
8 00:00:10,979 –> 00:00:12,630 32 we will talk about the
9 00:00:12,640 –> 00:00:14,369 intermediate value theorem
10 00:00:14,850 –> 00:00:15,250 there.
11 00:00:15,260 –> 00:00:16,709 As before what we will need
12 00:00:16,719 –> 00:00:18,209 is a continuous function.
13 00:00:18,219 –> 00:00:20,079 As usual
14 00:00:20,090 –> 00:00:21,170 we call the domain of the
15 00:00:21,180 –> 00:00:23,170 function here just by I.
16 00:00:23,559 –> 00:00:24,850 However, now
17 00:00:24,860 –> 00:00:26,270 actually we need an
18 00:00:26,280 –> 00:00:26,930 interval.
19 00:00:27,479 –> 00:00:29,340 So let’s write it as A B
20 00:00:29,350 –> 00:00:31,299 where A and B are included
21 00:00:31,309 –> 00:00:32,098 in the domain.
22 00:00:32,139 –> 00:00:33,779 In this case, we can draw
23 00:00:33,790 –> 00:00:35,450 a very nice compact picture
24 00:00:35,459 –> 00:00:36,909 for the graph of the function.
25 00:00:37,220 –> 00:00:38,729 For example, the graph could
26 00:00:38,740 –> 00:00:40,159 look like this where we have
27 00:00:40,169 –> 00:00:41,970 the interval A B on the X
28 00:00:41,979 –> 00:00:42,299 axis.
29 00:00:43,139 –> 00:00:44,450 And now you should see on
30 00:00:44,459 –> 00:00:46,200 the Y axis, we also
31 00:00:46,209 –> 00:00:47,259 find an interval.
32 00:00:48,240 –> 00:00:49,750 In this case, we find F of
33 00:00:49,759 –> 00:00:51,459 A at the bottom and F of
34 00:00:51,470 –> 00:00:52,459 B at the top.
35 00:00:52,880 –> 00:00:54,180 However, of course, it could
36 00:00:54,189 –> 00:00:55,419 be the other way around.
37 00:00:56,119 –> 00:00:57,590 Now, the intermediate value
38 00:00:57,599 –> 00:00:59,250 theorem tells us that for
39 00:00:59,259 –> 00:01:00,919 all values Y in this
40 00:01:00,930 –> 00:01:02,279 interval, we find a
41 00:01:02,290 –> 00:01:03,599 corresponding X.
42 00:01:04,180 –> 00:01:06,160 So you could say all intermediate
43 00:01:06,169 –> 00:01:07,669 values are hit by the
44 00:01:07,680 –> 00:01:09,339 function, the function
45 00:01:09,349 –> 00:01:11,029 simply cannot escape here
46 00:01:11,040 –> 00:01:12,650 if it starts here and ends
47 00:01:12,660 –> 00:01:14,220 there because it’s
48 00:01:14,230 –> 00:01:15,879 continuous and it cannot
49 00:01:15,889 –> 00:01:17,139 make any jumps.
50 00:01:17,940 –> 00:01:19,449 Therefore, the intermediate
51 00:01:19,459 –> 00:01:21,300 value theorem gives a meaning
52 00:01:21,309 –> 00:01:22,389 to the sentence that the
53 00:01:22,400 –> 00:01:23,889 graph of a continuous
54 00:01:23,900 –> 00:01:25,839 function can be drawn in
55 00:01:25,849 –> 00:01:26,680 one stroke.
56 00:01:27,150 –> 00:01:27,510 OK.
57 00:01:27,519 –> 00:01:29,040 Then let’s formulate this
58 00:01:29,050 –> 00:01:30,230 nice theorem
59 00:01:31,230 –> 00:01:31,540 here.
60 00:01:31,550 –> 00:01:33,190 I can tell you the intermediate
61 00:01:33,199 –> 00:01:34,709 value theorem is one of the
62 00:01:34,720 –> 00:01:36,290 most important theorems you
63 00:01:36,300 –> 00:01:37,709 learn in analysis.
64 00:01:38,160 –> 00:01:39,569 Now what we have to put in,
65 00:01:39,580 –> 00:01:41,339 we already know we
66 00:01:41,349 –> 00:01:42,769 only need two assumptions.
67 00:01:42,779 –> 00:01:44,330 Firstly, F should be a
68 00:01:44,339 –> 00:01:46,330 continuous function defined
69 00:01:46,339 –> 00:01:47,879 on the interval A B.
70 00:01:48,400 –> 00:01:49,809 And secondly, we take any
71 00:01:49,819 –> 00:01:51,589 number Y on the Y axis in
72 00:01:51,599 –> 00:01:53,080 this interval.
73 00:01:53,639 –> 00:01:53,980 OK.
74 00:01:53,989 –> 00:01:55,529 If we want to be precise,
75 00:01:55,540 –> 00:01:57,300 we write down both intervals.
76 00:01:57,309 –> 00:01:58,389 The one where we start with
77 00:01:58,400 –> 00:01:59,919 F of A and the other one
78 00:01:59,930 –> 00:02:00,919 where we start with F of
79 00:02:00,930 –> 00:02:02,900 B just depending which
80 00:02:02,910 –> 00:02:04,110 of the two numbers is the
81 00:02:04,120 –> 00:02:05,059 smaller one.
82 00:02:05,449 –> 00:02:07,059 For example, in our picture,
83 00:02:07,069 –> 00:02:08,679 we can set y to this
84 00:02:08,690 –> 00:02:09,300 point.
85 00:02:09,750 –> 00:02:10,979 And then the intermediate
86 00:02:10,990 –> 00:02:12,839 value theorem tells us there
87 00:02:12,850 –> 00:02:14,559 is a corresponding X, we
88 00:02:14,570 –> 00:02:15,789 call X tilde.
89 00:02:16,419 –> 00:02:17,880 It’s an element of the interval
90 00:02:17,889 –> 00:02:19,690 A B and it is sent to
91 00:02:19,699 –> 00:02:20,960 Y by F.
92 00:02:21,580 –> 00:02:23,309 So F of X tilde is
93 00:02:23,320 –> 00:02:24,309 equal to Y.
94 00:02:25,000 –> 00:02:25,399 OK.
95 00:02:25,410 –> 00:02:26,820 So this is the intermediate
96 00:02:26,830 –> 00:02:28,399 value theorem we’ll prove
97 00:02:28,410 –> 00:02:28,880 today.
98 00:02:29,460 –> 00:02:30,639 But before we do that, I
99 00:02:30,649 –> 00:02:32,619 present you corollary you really
100 00:02:32,630 –> 00:02:33,520 should remember.
101 00:02:34,240 –> 00:02:36,179 It tells us that the image
102 00:02:36,190 –> 00:02:37,899 of the interval A B
103 00:02:37,910 –> 00:02:39,289 where we denote the image
104 00:02:39,300 –> 00:02:40,360 with the square brackets as
105 00:02:40,369 –> 00:02:42,179 well is also an
106 00:02:42,190 –> 00:02:42,899 interval.
107 00:02:43,360 –> 00:02:44,619 More concretely it’s the
108 00:02:44,630 –> 00:02:45,979 interval that starts with
109 00:02:45,990 –> 00:02:47,289 the minimum of the function
110 00:02:47,300 –> 00:02:49,220 F and goes to the maximum
111 00:02:49,229 –> 00:02:50,139 of the function F.
112 00:02:50,940 –> 00:02:52,070 Therefore, you see it’s the
113 00:02:52,080 –> 00:02:53,440 same claim as before.
114 00:02:53,490 –> 00:02:55,179 All the intermediate values
115 00:02:55,190 –> 00:02:56,000 are also hit.
116 00:02:56,710 –> 00:02:57,070 OK.
117 00:02:57,080 –> 00:02:58,729 Then I would say we can start
118 00:02:58,740 –> 00:03:00,440 with the proof of the intermediate
119 00:03:00,449 –> 00:03:01,580 value theorem.
120 00:03:02,210 –> 00:03:03,990 In fact, as we will see the
121 00:03:04,000 –> 00:03:05,869 proof is not so complicated,
122 00:03:06,529 –> 00:03:07,910 the first step will be to
123 00:03:07,919 –> 00:03:09,449 normalize the problem in
124 00:03:09,460 –> 00:03:10,229 some sense.
125 00:03:10,860 –> 00:03:12,130 In order to understand this,
126 00:03:12,139 –> 00:03:13,500 let’s visualize the graph
127 00:03:13,580 –> 00:03:14,710 of a function again,
128 00:03:15,440 –> 00:03:17,029 this is similar as before
129 00:03:17,080 –> 00:03:18,419 where now, for example, we
130 00:03:18,429 –> 00:03:19,710 have our Y here.
131 00:03:20,490 –> 00:03:21,669 Now when I say we want to
132 00:03:21,679 –> 00:03:23,330 normalize the problem, it
133 00:03:23,339 –> 00:03:24,990 means it would be nice to
134 00:03:25,000 –> 00:03:26,910 have this Y at zero.
135 00:03:27,479 –> 00:03:28,729 And of course, we could just
136 00:03:28,740 –> 00:03:29,289 do that.
137 00:03:29,300 –> 00:03:31,089 So we take the whole function
138 00:03:31,149 –> 00:03:32,710 and shift it to zero.
139 00:03:33,520 –> 00:03:34,960 So simply as that we get
140 00:03:34,970 –> 00:03:36,339 a new function, we could
141 00:03:36,350 –> 00:03:37,229 call G.
142 00:03:38,050 –> 00:03:38,960 Now, what you should see
143 00:03:38,970 –> 00:03:40,960 is that the problem for searching
144 00:03:40,970 –> 00:03:42,580 such an x tilde stays
145 00:03:42,589 –> 00:03:43,809 exactly the same.
146 00:03:43,820 –> 00:03:45,250 But now we search for
147 00:03:45,259 –> 00:03:45,910 zeros.
148 00:03:46,660 –> 00:03:48,470 In summary, this is our idea,
149 00:03:48,479 –> 00:03:49,899 we simply define a new
150 00:03:49,910 –> 00:03:51,770 function G and
151 00:03:51,779 –> 00:03:53,630 this is simply F minus
152 00:03:53,639 –> 00:03:54,830 the value Y.
153 00:03:55,529 –> 00:03:56,880 Now, the next normalization
154 00:03:56,889 –> 00:03:58,360 we could do is that it would
155 00:03:58,369 –> 00:04:00,190 be nice if the value at the
156 00:04:00,199 –> 00:04:02,169 right is larger than
157 00:04:02,179 –> 00:04:03,550 the value on the left.
158 00:04:04,029 –> 00:04:05,429 So what we could do is just
159 00:04:05,440 –> 00:04:06,830 mirror the whole graph.
160 00:04:06,860 –> 00:04:08,339 And then we have a new
161 00:04:08,350 –> 00:04:10,169 function which has exactly
162 00:04:10,179 –> 00:04:12,169 this property and
163 00:04:12,179 –> 00:04:13,649 this function we now can
164 00:04:13,660 –> 00:04:15,009 call F tilde.
165 00:04:15,479 –> 00:04:17,130 However, please note here,
166 00:04:17,140 –> 00:04:18,890 we only mirror the graph
167 00:04:18,899 –> 00:04:19,970 if it’s needed.
168 00:04:20,450 –> 00:04:21,890 So in the formula, this would
169 00:04:21,899 –> 00:04:23,690 mean F tilde is defined
170 00:04:23,700 –> 00:04:25,089 as minus G.
171 00:04:25,500 –> 00:04:27,040 However, only in the case
172 00:04:27,049 –> 00:04:28,970 that G of A is greater than
173 00:04:28,980 –> 00:04:29,570 zero.
174 00:04:30,019 –> 00:04:31,920 If G of A is less or equal
175 00:04:31,929 –> 00:04:33,109 than zero, we don’t have
176 00:04:33,119 –> 00:04:34,130 to do anything.
177 00:04:34,140 –> 00:04:35,980 We can just set F tilde to
178 00:04:35,989 –> 00:04:36,399 G.
179 00:04:37,049 –> 00:04:37,429 OK.
180 00:04:37,440 –> 00:04:38,950 Now let’s summarize what
181 00:04:38,959 –> 00:04:39,790 we have here.
182 00:04:40,640 –> 00:04:42,420 First F tilde is of
183 00:04:42,429 –> 00:04:44,130 course, still a continuous
184 00:04:44,140 –> 00:04:44,869 function.
185 00:04:45,290 –> 00:04:47,269 And our value we search for,
186 00:04:47,279 –> 00:04:49,250 we could call why tilde is
187 00:04:49,260 –> 00:04:50,429 just zero.
188 00:04:51,059 –> 00:04:52,709 Moreover, we also know that
189 00:04:52,720 –> 00:04:54,660 we have a minus sign or zero
190 00:04:54,670 –> 00:04:56,420 on the left hand side and
191 00:04:56,429 –> 00:04:58,160 the plus sign or zero on
192 00:04:58,170 –> 00:04:59,179 the right hand side.
193 00:04:59,850 –> 00:05:01,239 Therefore, the f tilde we
194 00:05:01,250 –> 00:05:02,829 have drawn before is the
195 00:05:02,839 –> 00:05:04,170 correct example we should
196 00:05:04,179 –> 00:05:04,940 have in mind.
197 00:05:05,489 –> 00:05:05,839 OK.
198 00:05:05,850 –> 00:05:07,399 Now, the proof that we follow
199 00:05:07,410 –> 00:05:08,760 will look similar to the
200 00:05:08,769 –> 00:05:10,179 proof we already did for
201 00:05:10,190 –> 00:05:12,019 the Bolzano Weierstrass theorem,
202 00:05:12,730 –> 00:05:14,549 namely we cut the interval
203 00:05:14,559 –> 00:05:16,200 A B in the middle
204 00:05:16,480 –> 00:05:17,739 and then choose one of the
205 00:05:17,750 –> 00:05:18,760 two intervals here.
206 00:05:19,279 –> 00:05:20,630 And after this, this
207 00:05:20,640 –> 00:05:21,920 bisection, we will do
208 00:05:21,929 –> 00:05:22,799 repeatedly.
209 00:05:23,619 –> 00:05:24,989 In conclusion in the end,
210 00:05:25,000 –> 00:05:26,609 we will get a limit here.
211 00:05:27,380 –> 00:05:28,470 The only thing we have to
212 00:05:28,480 –> 00:05:30,190 do in each step is to check
213 00:05:30,200 –> 00:05:32,119 the value F tile of C
214 00:05:32,750 –> 00:05:34,489 if this one is greater than
215 00:05:34,500 –> 00:05:35,089 zero.
216 00:05:35,109 –> 00:05:37,089 Then our C could be the new
217 00:05:37,100 –> 00:05:37,529 B.
218 00:05:38,390 –> 00:05:39,779 In other words, we will choose
219 00:05:39,790 –> 00:05:40,940 the left interval.
220 00:05:40,950 –> 00:05:42,779 In this case. In the other
221 00:05:42,790 –> 00:05:43,950 case, we would choose the
222 00:05:43,959 –> 00:05:45,029 right interval.
223 00:05:45,559 –> 00:05:47,109 Of course, the overall idea
224 00:05:47,119 –> 00:05:48,350 of this definition is that
225 00:05:48,359 –> 00:05:50,149 we still have the same assumption,
226 00:05:50,160 –> 00:05:51,989 we still have the sign change
227 00:05:52,000 –> 00:05:53,070 from left to right.
228 00:05:53,809 –> 00:05:55,230 This means then that nothing
229 00:05:55,239 –> 00:05:56,660 really changes here, we can
230 00:05:56,670 –> 00:05:57,899 do the same thing again.
231 00:05:58,630 –> 00:05:59,640 Then of course, we would
232 00:05:59,649 –> 00:06:01,529 define B two and A two.
233 00:06:01,540 –> 00:06:02,880 Then afterwards we would
234 00:06:02,890 –> 00:06:04,549 define B three and A three
235 00:06:04,559 –> 00:06:05,309 and so on.
236 00:06:05,880 –> 00:06:07,410 Therefore, in the end, we
237 00:06:07,420 –> 00:06:09,049 get two sequences out.
238 00:06:09,459 –> 00:06:11,429 So we have A and BN.
239 00:06:11,440 –> 00:06:12,959 And by construction, we know
240 00:06:12,970 –> 00:06:14,640 they are called Cauchy sequences.
241 00:06:15,339 –> 00:06:16,940 Moreover, we also know that
242 00:06:16,950 –> 00:06:18,220 the intervals get smaller
243 00:06:18,230 –> 00:06:19,049 and smaller.
244 00:06:19,059 –> 00:06:20,260 And indeed, we know that
245 00:06:20,269 –> 00:06:22,119 the length converges to zero,
246 00:06:22,880 –> 00:06:24,790 hence the completeness axiom
247 00:06:24,799 –> 00:06:26,250 and the limit theorems get
248 00:06:26,260 –> 00:06:27,510 us to the result that we
249 00:06:27,519 –> 00:06:29,429 have exactly one limit here.
250 00:06:29,799 –> 00:06:31,149 Now, you might already know
251 00:06:31,160 –> 00:06:32,570 this is exactly the point
252 00:06:32,579 –> 00:06:34,109 we want to call X tilde.
253 00:06:34,200 –> 00:06:35,640 And by construction it’s
254 00:06:35,649 –> 00:06:37,149 an element of the interval
255 00:06:37,160 –> 00:06:37,809 A B.
256 00:06:38,470 –> 00:06:39,890 Hence the only question that
257 00:06:39,899 –> 00:06:41,649 remains is what happens with
258 00:06:41,660 –> 00:06:43,359 X tilde when we put it into
259 00:06:43,369 –> 00:06:44,690 the function F tilde.
260 00:06:45,489 –> 00:06:46,809 In order to see this, please
261 00:06:46,820 –> 00:06:48,750 note that the sign change
262 00:06:48,760 –> 00:06:49,929 for the function F tilde
263 00:06:49,940 –> 00:06:51,899 holds for the whole sequence
264 00:06:51,910 –> 00:06:53,329 A and BN.
265 00:06:54,119 –> 00:06:56,019 Therefore, and by the monotonicity
266 00:06:56,029 –> 00:06:57,820 of the limit, we also know
267 00:06:57,829 –> 00:06:59,279 that this property still
268 00:06:59,290 –> 00:07:00,839 holds when we put the limit
269 00:07:00,850 –> 00:07:01,410 in front.
270 00:07:01,959 –> 00:07:03,339 And there you see, we can
271 00:07:03,350 –> 00:07:04,970 use the continuity of the
272 00:07:04,980 –> 00:07:06,760 function of tilde and push
273 00:07:06,769 –> 00:07:07,859 the limit inside.
274 00:07:08,519 –> 00:07:09,950 Then you see, it looks like
275 00:07:09,959 –> 00:07:11,670 this and we can substitute
276 00:07:11,679 –> 00:07:13,250 this limit with X tilde.
277 00:07:13,899 –> 00:07:15,790 So what we get out is a number
278 00:07:15,799 –> 00:07:17,790 F tilt of X tilde which
279 00:07:17,799 –> 00:07:19,429 is less or equal than zero
280 00:07:19,440 –> 00:07:20,720 and greater or equal than
281 00:07:20,730 –> 00:07:21,220 zero.
282 00:07:21,779 –> 00:07:23,489 Hence, the only possibility
283 00:07:23,500 –> 00:07:25,100 is that this number is
284 00:07:25,109 –> 00:07:26,399 exactly zero.
285 00:07:26,890 –> 00:07:28,420 And there you see indeed
286 00:07:28,429 –> 00:07:29,790 we have solved the problem
287 00:07:29,799 –> 00:07:31,500 for our function F tilde.
288 00:07:32,170 –> 00:07:33,359 Also, you should see it’s
289 00:07:33,369 –> 00:07:34,839 not a problem at all to go
290 00:07:34,850 –> 00:07:36,649 back to our original function
291 00:07:36,660 –> 00:07:38,359 F because you
292 00:07:38,369 –> 00:07:40,200 immediately see the X tilde
293 00:07:40,209 –> 00:07:41,730 we found is also a
294 00:07:41,739 –> 00:07:43,420 zero for the function G.
295 00:07:43,940 –> 00:07:45,079 This simply holds by the
296 00:07:45,089 –> 00:07:46,279 definition of the function
297 00:07:46,290 –> 00:07:48,079 F tilde as a flip when we
298 00:07:48,089 –> 00:07:48,559 need it.
299 00:07:49,290 –> 00:07:51,200 However, the function G was
300 00:07:51,209 –> 00:07:52,910 defined by F minus
301 00:07:52,950 –> 00:07:54,839 Y and because this
302 00:07:54,850 –> 00:07:56,220 is now zero, this
303 00:07:56,230 –> 00:07:58,109 implies F
304 00:07:58,119 –> 00:07:59,130 of X tilde is
305 00:07:59,140 –> 00:08:00,779 indeed Y
306 00:08:01,369 –> 00:08:02,570 and that’s exactly what we
307 00:08:02,579 –> 00:08:03,359 wanted to show.
308 00:08:03,410 –> 00:08:04,809 So the intermediate value
309 00:08:04,820 –> 00:08:06,279 theorem is proven.
310 00:08:06,920 –> 00:08:07,380 OK.
311 00:08:07,390 –> 00:08:09,209 Then I would say please remember
312 00:08:09,220 –> 00:08:10,709 this important theorem and
313 00:08:10,720 –> 00:08:11,980 then I see you in the next
314 00:08:11,989 –> 00:08:12,440 video.
315 00:08:12,970 –> 00:08:14,820 Have a nice day and bye.
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Quiz Content
Q1: What is a correct formulation for the intermediate value theorem?
A1: Each continuous function $f: [a,b] \rightarrow \mathbb{R}$ has a maximum and a minimum.
A2: For each continuous function $f: [a,b] \rightarrow \mathbb{R}$ and each element $y$ between $f(a)$ and $f(b)$, there is a point $x \in [a,b]$ with $f(x) = y$.
A3: For each bounded function $f: [a,b] \rightarrow \mathbb{R}$ and each $y$ between $f(a)$ and $f(b)$, there is a point $x \in [a,b]$ with $f(x) = y$.
A4: For each continuous function $f: [a,b] \rightarrow \mathbb{R}$ and each element $x \in [a,b]$, there is an element $y$ with $f(x) = y$.
A5: For each bounded function $f: [a,b] \rightarrow \mathbb{R}$ and each element $x \in [a,b]$, there is an element $y$ with $f(x) = y$.
Q2: Which statement for a continuous function $f: [a,b] \rightarrow \mathbb{R}$ is, in general, not correct?
A1: If $f(a)=-1$ and $f(b) = 1$, then $f$ has at least one zero.
A2: If $f(a) = f(b)$, then $f$ is constant.
A3: The image $f\Big[ [a,b] \Big]$ is an interval.
A4: If $f(a)=-1$ and $f(b) = 1$, then there is a point $x$ with $f(x) = 0.5$.
Q3: Is there an example of a continuous function $f: [-1,1] \rightarrow \mathbb{R}$ with $f(-1) = 0$ and $f(1) = 2$ where also $f(0) = -4$.
A1: Yes!
A2: No!
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Last update: 2025-01
