Advanced Calculus (I)

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Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

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4.4 Monotone Function and The Inverse Function Theorem

Definition

Let E be a nonempty subset ofRandf :E →R.

(i)

f is said to beincreasing(respectively,strictly increasing) on E if and only ifx1,x2 ∈E andx1<x2imply

f(x1)≤f(x2)(respectively, f(x1)<f(x2)).

(ii)

f is said to bedereasing(respectively,strictly decreasing) on E if and only ifx1,x2 ∈E andx1<x2imply

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4.4 Monotone Function and The Inverse Function Theorem

Definition

(i)

(ii)

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4.4 Monotone Function and The Inverse Function Theorem

Definition

(i)

(ii)

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4.4 Monotone Function and The Inverse Function Theorem

Definition

(i)

(ii)

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4.4 Monotone Function and The Inverse Function Theorem

Definition

(i)

(ii)

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4.4 Monotone Function and The Inverse Function Theorem

Definition

(i)

(ii)

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Definition (iii)

f is said to bemonotone(respectively,strictly monotone) on E if and only if f is either decreasing or increasing (respectively, either strictly decreasing or strictly increasing) on E.

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Definition (iii)

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Definition (iii)

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Theorem

Suppose that a,b ∈R, with a6=b, that f is continuous on [a,b], and that f is differentiable on (a,b).

(i)

If f⁰(x)>0(respectively, f’(x)¡0) for all x ∈(a,b), then f is strictly increasing (respectively, strictly decreasing) on [a,b].

(ii)

If f⁰(x) =0for all x ∈(a,b), then f is constant on [a,b].

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Theorem

(i)

(ii)

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Theorem

(i)

(ii)

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Theorem

(i)

(ii)

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Theorem

(i)

(ii)

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Theorem

(i)

(ii)

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Proof:

Leta≤x1 <x2 ≤b. By the Mean Value Theorem,there is ac ∈(a,b)such thatf(x2)−f(x1) =f⁰(c)(x2−x1).

Thus,f(x2)>f(x1)whenf⁰(c)>0 andf(x2)<f(x1)when f⁰(c)<0.This proves part (i).

To prove part (ii), leta≤x ≤b.By the Mean Value Theorem and hypothesis there is ac ∈(a,b)such that

f(x)−f(a) =f⁰(c)(x −a) =0.

Thus,f(x) =f(a)for allx ∈[a,b]. 2

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Proof:

Leta≤x1 <x2 ≤b. By the Mean Value Theorem, there is ac ∈(a,b)such thatf(x2)−f(x1) =f⁰(c)(x2−x1).

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

Leta≤x1 <x2 ≤b. By the Mean Value Theorem,there is ac ∈(a,b)such thatf(x2)−f(x1) =f⁰(c)(x2−x1).

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

To prove part (ii),leta≤x ≤b.By the Mean Value Theorem and hypothesis there is ac ∈(a,b)such that

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

To prove part (ii),leta≤x ≤b.By the Mean Value Theorem and hypothesis there is ac ∈(a,b)such that

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Proof:

f(x)−f(a) =f⁰(c)(x −a) =0.

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Theorem

If f is 1-1 and continuous on an interval I, then f is strictly monotone on I and f⁻¹ is continuous and strictly

monotone on f(I).

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Theorem

If f is 1-1 and continuous on an interval I, then f is strictly monotone on I and f⁻¹ is continuous and strictly

monotone on f(I).

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Theorem (Inverse Function Theorem)

Let f be 1-1 and continuous on an open interval I. If a∈f(I)and if f⁰(f⁻¹(a))exists and is nonzero, then f⁻¹ is differentiable at a and

(f⁻¹)⁰(a) = 1 f⁰(f⁻¹(a)).

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Theorem (Inverse Function Theorem)

Let f be 1-1 and continuous on an open interval I. If a∈f(I)and if f⁰(f⁻¹(a))exists and is nonzero, then f⁻¹ is differentiable at a and

(f⁻¹)⁰(a) = 1 f⁰(f⁻¹(a)).

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Lemma:

Suppose that f is increasing on [a,b].

(i)Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii) Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

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Lemma:

(i) Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

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Lemma:

(i)Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii)Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

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Lemma:

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Lemma:

(ii)Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

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Lemma:

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Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E.Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

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Proof:

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence,s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

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Proof:

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E.Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

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Proof:

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence,s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Proof:

s− <f(x1)≤f(x)≤s

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Theorem

If f is monotone on an open interval I, then f has at most countably many points of discontinuity on I.

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Theorem

If f is monotone on an open interval I, then f has at most countably many points of discontinuity on I.

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Theorem (Intermediate Value Theorem For Derivatives) Suppose that f is differentiable on [a,b] with f⁰(a)6=f⁰(b).

If y0 is a real number that lies between f⁰(a)and f⁰(b), then there is an x0∈(a,b)such that f⁰(x0) =y0.

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Theorem (Intermediate Value Theorem For Derivatives) Suppose that f is differentiable on [a,b] with f⁰(a)6=f⁰(b).

If y0 is a real number that lies between f⁰(a)and f⁰(b), then there is an x0∈(a,b)such that f⁰(x0) =y0.

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Advanced Calculus (I)