The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the...
Transcript of The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the...
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Why the Intermediate Value Theorem may be true
Statement of the Intermediate Value Theorem
Reduction to the Special Case where f (a) < f (b)
Reduction to the Special Case where γ = 0
Special Case of the Intermediate Value Theorem
Proof: Definition of SCase 1. g (c) < 0Case 2. g (c) > 0In Conclusion.
![Page 3: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/3.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 4: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/4.jpg)
Why the Intermediate Value Theorem may be true
We start with a
closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 5: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/5.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].
We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 6: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/6.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 7: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/7.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).
From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 8: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/8.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x).
Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 9: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/9.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for which
f (c) = γ.
![Page 10: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/10.jpg)
Why the Intermediate Value Theorem may be true
We start with a closedinterval [a, b].We next take a function fcontinuous on [a, b].
Then we take any value γlying between f (a) andf (b).From the graph it doesn’tseem unreasonable thatthe line y = γ intersectsthe curve y = f (x). Inwhich case there existsc : a < c < b for whichf (c) = γ.
![Page 11: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/11.jpg)
Statement of the Result
Theorem (Bolzano 1817. Intermediate Value Theorem)
Suppose that f is a function continuous on a closed interval [a, b]and that f (a) 6= f (b). If γ is some number between f (a) andf (b) then there must be at least one c : a < c < b for whichf (c) = γ.
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Statement of the Result
Theorem (Bolzano 1817. Intermediate Value Theorem)
Suppose that f is a function continuous on a closed interval [a, b]and that f (a) 6= f (b). If γ is some number between f (a) andf (b) then there must be at least one c : a < c < b for whichf (c) = γ.
![Page 13: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/13.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 14: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/14.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).
So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 15: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/15.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)?
For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 16: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/16.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 17: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/17.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 18: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/18.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x)
for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 19: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/19.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 20: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/20.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e.
g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 21: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/21.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 22: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/22.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for which
g(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 23: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/23.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ
i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 24: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/24.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.
So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 25: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/25.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.
Thus we may assumef (a) < f (b).
![Page 26: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/26.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.
Thus we may assumef (a) < f (b).
![Page 27: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/27.jpg)
Reduction to the Special Case where f (a) < f (b)
We will prove the Theoremonly when f (a) < f (b).So how do we deal withf (a) > f (b)? For instance
Answer: we considerg(x) = −f (x) for whichg(a) < g(b).
Given γ : f (b) < γ < f (a)we have
−f (a) < −γ < −f (b),
i.e. g(a) < −γ < g(b).
For this g we can applythe Intermediate ValueTheorem to find ac : a < c < b for whichg(c) = −γ i.e.−f (c) = −γ.So we have again solvedf (c) = γ.Thus we may assumef (a) < f (b).
![Page 28: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/28.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 29: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/29.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.
By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 30: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/30.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).
So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 31: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/31.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0?
For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 32: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/32.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
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Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.
We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 34: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/34.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0,
i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 35: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/35.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.
So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 36: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/36.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.
Thus we may assumeγ = 0
![Page 37: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/37.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.
Thus we may assumeγ = 0
![Page 38: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/38.jpg)
Reduction to the Special Case where γ = 0
We will prove the Theoremonly in the case γ = 0.By the previous slide thismeans we will be assumingthat f (a) < 0 < f (b).So how do we deal withγ 6= 0? For instance
Answer: we considerg(x) = f (x)− γ.We will then be able toapply the Special Case ofthe intermediate ValueTheorem to find ac : a < c < b for whichg(c) = 0, i.e.f (c)− γ = 0.So we have again solvedf (c) = γ.Thus we may assumeγ = 0
![Page 39: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/39.jpg)
Intermediate Value Theorem
TheoremSuppose that g is a function continuous on a closed interval [a, b]and that g (a) < 0 < g (b). Then there must be at least onec : a < c < b for which g (c) = 0.
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Intermediate Value Theorem
TheoremSuppose that g is a function continuous on a closed interval [a, b]and that g (a) < 0 < g (b). Then there must be at least onec : a < c < b for which g (c) = 0.
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Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
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Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 43: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/43.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.
Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 44: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/44.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.
Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 45: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/45.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.
Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 46: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/46.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.
Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 47: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/47.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .
There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 48: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/48.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely
g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 49: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/49.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0,
g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 50: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/50.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0
org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 51: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/51.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0.
We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 52: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/52.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction.
Thus we will conclude that g (c) = 0 as required.
![Page 53: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/53.jpg)
Proof: Definition of S
Let g be a function continuous on a closed interval [a, b] withg (a) < 0 < g (b):
Let S = {x ∈ [a, b] : g (x) < 0}.Then g (a) < 0⇒ a ∈ S ⇒ S 6= ∅.Also all x ∈ S satisfies x ≤ b so S is bounded above.Therefore by the Completeness axiom of R we deduce that lubS exists.Write lubS =c .There are three possibilities for c namely g(c) < 0, g(c) > 0 org(c) = 0. We will show that each of the first two possibilities lead to acontradiction. Thus we will conclude that g (c) = 0 as required.
![Page 54: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/54.jpg)
Case 1. g (c) < 0
Assume g (c) < 0.
Note g (b) > 0 so c 6= b thus c ∈ [a, b).
The function g continuous on [a, b] means, in particular, that
limx→`+ g (x) = g (`) for all ` ∈ [a, b). Since c ∈ [a, b) we conclude that
limx→c+ g (x) = g (c).
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Case 1. g (c) < 0
Assume g (c) < 0.
Note g (b) > 0 so c 6= b thus c ∈ [a, b).
The function g continuous on [a, b] means, in particular, that
limx→`+ g (x) = g (`) for all ` ∈ [a, b). Since c ∈ [a, b) we conclude that
limx→c+ g (x) = g (c).
![Page 56: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/56.jpg)
Case 1. g (c) < 0
Assume g (c) < 0.
Note g (b) > 0 so c 6= b thus c ∈ [a, b).
The function g continuous on [a, b] means, in particular, that
limx→`+ g (x) = g (`) for all ` ∈ [a, b). Since c ∈ [a, b) we conclude that
limx→c+ g (x) = g (c).
![Page 57: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/57.jpg)
Case 1. g (c) < 0
Assume g (c) < 0.
Note g (b) > 0 so c 6= b thus c ∈ [a, b).
The function g continuous on [a, b] means, in particular, that
limx→`+ g (x) = g (`) for all ` ∈ [a, b). Since c ∈ [a, b) we conclude that
limx→c+ g (x) = g (c).
![Page 58: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/58.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 59: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/59.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown.
Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 60: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/60.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+
tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 61: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/61.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ then
g(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 62: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/62.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 63: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/63.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 64: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/64.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.
For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 65: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/65.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c
and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 66: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/66.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 67: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/67.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ S
which implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 68: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/68.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .
Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 69: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/69.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c ,
which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 70: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/70.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0,
acontradiction.
![Page 71: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/71.jpg)
Case 1. g (c) < 0 and limx→c+ g (x) = g (c)
We look to the right of c asshown. Use ε = −g (c) /2 > 0in the definition of limx→c+ tofind δ > 0 such that ifc < x < c + δ theng(c)− ε < g(x) < g(c) + ε.
In particular,g(x) < g(c)− g(c)/2= g(c)/2 < 0.
Pick any one of these x so, forinstance, choose x0 = c + δ/2.For this value we have bothx0 > c and g(x0) < 0.
But g(x0) < 0 implies x0 ∈ Swhich implies x0 ≤ lubS = c .Thus we have both x0 > cand x0 ≤ c , which combine asx0 > c ≥ x0, i.e. x0 > x0, acontradiction.
![Page 72: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/72.jpg)
Case 2. g (c) > 0
Since g (a) < 0 we have c 6= a and thus c ∈ (a, b].
The function g continuous on [a, b] means, in particular, thatlimx→`− g (x) = g (`) for all ` ∈ (a, b].
Since c ∈ (a, b] we conclude that limx→c− g (x) = g (c).
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Case 2. g (c) > 0
Since g (a) < 0 we have c 6= a and thus c ∈ (a, b].
The function g continuous on [a, b] means, in particular, thatlimx→`− g (x) = g (`) for all ` ∈ (a, b].
Since c ∈ (a, b] we conclude that limx→c− g (x) = g (c).
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Case 2. g (c) > 0
Since g (a) < 0 we have c 6= a and thus c ∈ (a, b].
The function g continuous on [a, b] means, in particular, thatlimx→`− g (x) = g (`) for all ` ∈ (a, b].
Since c ∈ (a, b] we conclude that limx→c− g (x) = g (c).
![Page 75: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/75.jpg)
Case 2. g (c) > 0
Since g (a) < 0 we have c 6= a and thus c ∈ (a, b].
The function g continuous on [a, b] means, in particular, thatlimx→`− g (x) = g (`) for all ` ∈ (a, b].
Since c ∈ (a, b] we conclude that limx→c− g (x) = g (c).
![Page 76: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/76.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 77: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/77.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown.
Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 78: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/78.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c−
tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 79: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/79.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c then
g(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 80: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/80.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε.
Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 81: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/81.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 82: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/82.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)
But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 83: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/83.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S.
Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 84: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/84.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case
thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 85: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/85.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.
But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 86: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/86.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.
So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 87: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/87.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, a
contradiction.
![Page 88: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/88.jpg)
Case 2. g (c) > 0 and limx→c− g (x) = g (c)
We look to the left of c asshown. Use ε = g (c) /2 > 0in the definition of limx→c− tofind δ > 0 such that ifc − δ < x < c theng(c) + ε > g(x) > g(c)− ε. Inparticular, g(x) > g(c)/2 > 0.
Hence, if c − δ < x < c theng(x) > 0. (1)But, by definition, c is theleast upper bound for S. Thusc − δ is not an upper boundfor S, in which case thereexists x1 with c − δ < x1 ≤ cand x1 ∈ S i.e. g (x1) < 0.But from (1) we haveg (x1) > 0.So we have both g (x1) < 0and g (x1) > 0, acontradiction.
![Page 89: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/89.jpg)
In Conclusion.
We definedc = lub {x ∈ [a, b] : g (x) < 0} .
It must be the case that one of the following hold:
g(c) < 0, g(c) > 0 or g(c) = 0.
We then showed that both g(c) < 0 and g(c) > 0 lead to contradictions.Hence g(c) = 0 as required.
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![Page 90: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/90.jpg)
In Conclusion.
We definedc = lub {x ∈ [a, b] : g (x) < 0} .
It must be the case that one of the following hold:
g(c) < 0, g(c) > 0 or g(c) = 0.
We then showed that both g(c) < 0 and g(c) > 0 lead to contradictions.Hence g(c) = 0 as required.
�
![Page 91: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/91.jpg)
In Conclusion.
We definedc = lub {x ∈ [a, b] : g (x) < 0} .
It must be the case that one of the following hold:
g(c) < 0, g(c) > 0 or g(c) = 0.
We then showed that both g(c) < 0 and g(c) > 0 lead to contradictions.Hence g(c) = 0 as required.
�
![Page 92: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/92.jpg)
In Conclusion.
We definedc = lub {x ∈ [a, b] : g (x) < 0} .
It must be the case that one of the following hold:
g(c) < 0, g(c) > 0 or g(c) = 0.
We then showed that both g(c) < 0 and g(c) > 0 lead to contradictions.
Hence g(c) = 0 as required.
�
![Page 93: The Intermediate Value Theorem - University of …mdc/MATH20101/notesPermanant/IVTheorem.pdfWhy the Intermediate Value Theorem may be true We start with a closed interval [a;b]. We](https://reader031.fdocuments.us/reader031/viewer/2022030421/5aa83d757f8b9a77188b54c4/html5/thumbnails/93.jpg)
In Conclusion.
We definedc = lub {x ∈ [a, b] : g (x) < 0} .
It must be the case that one of the following hold:
g(c) < 0, g(c) > 0 or g(c) = 0.
We then showed that both g(c) < 0 and g(c) > 0 lead to contradictions.Hence g(c) = 0 as required.
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