Intermediate Value Theorem
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Transcript of Intermediate Value Theorem
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Intermediate Value Theorem
2.4 cont.
![Page 2: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/2.jpg)
![Page 3: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/3.jpg)
Examples
If between 7am and 2pm the temperature went from 55 to 70. At some time it reached 62. Time is continuous
If between his 14th and 15th birthday, a boy went from 150 to 165 lbs. At some point he weighed 155lbs. It may have occurred more than once.
![Page 4: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/4.jpg)
Show that a “c” exists such that f(c)=2 for f(c)=x^2 +2x-3 in the interval [0, 2]f(x) is continuous on the interval
f(0)= -3
f(2)= 5
523- since2)("" IVTby cfca
![Page 5: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/5.jpg)
Determine if f(x) has any real roots [1,2] intervalin 1)( 2 xxxf
f(x) is continuous on the interval
f(1)= - #
f(2)= + #
0)(2,1 IVTby cfc
![Page 6: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/6.jpg)
Is any real number exactly one less than its cube?
(Note that this doesn’t ask what the number is, only if it exists.)
3 1x x
30 1x x
3 1f x x x
1 1f 2 5f
Since f is a continuous function, by the intermediate value theorem it must take on every value between -1 and 5.Therefore there must be at least one solution between 1 and 2.
Use your calculator to find an approximate solution.
1.32472
1)0( f
2
-2
-5 5
![Page 7: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/7.jpg)
Max-Min Theorem for Continuous Functions
If f is continuous at every point of the closed interval [a, b], then f takes on a minimum value m and a maximum value M on [a, b]. That is, there are numbers α and β in [a, b] such that f(α) = m, f(β) = M, and m ≤ f(x) ≤ M at all points x in [a, b]
Max and mins at interior points Max and mins at
endpoints
Min at interior point and max at endpoint
![Page 8: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/8.jpg)
Why does the IVT fail to hold for f(x) on [-1, 1]?
101
011)(
x
xxf
Not Continuous in interval!
Point of discontinuity at x = 0
1
-1
-2 2
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Show why a root exists in the given interval
]1,2[22)( 23 onxxxf
Continuous in interval
f(-2)= -10
f(-1)= 1
0)(1,2 IVTby cfc
![Page 10: Intermediate Value Theorem](https://reader036.fdocuments.net/reader036/viewer/2022080916/56812de4550346895d933c72/html5/thumbnails/10.jpg)
Show why a root exists in the given interval
1,211
1 4 onxx
Continuous in interval
f(1)= 1
0)(1,2
1 IVTby
cfc
16
1521
16
1)2
1(
f