The Intermediate Value Theorem

Dr M Colemancoleman@manchester.ac.uk

March 2009

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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

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

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

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

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

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

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

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

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

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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).

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).

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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).

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).

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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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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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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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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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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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