Chapter1Functions and LimitsS1.10Properties of Continuous Functions
Chapter 1 Functions and Limits §1.10 Properties of Continuous Functions
I. Max-Min Theorem1.Definion:Let I, the domain of f (x), contain the point xo.f(x, )is a maximum value of f(x)on I if f(x.)≥ f(x)for all x in I; f(x)is a minimum value of f(x)on I if f(x)≤ f(x)for all x in I;2. TheoremIf f (x) is continuous on a closed interval [a,b], then f (x)attains both a maximum and a minimum value there.If f(x) is continuous on a closed interval [a,bl, thenis f(x) bounded on [a,b].S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions I. Max-Min Theorem 1.Definion: Let I, the domain of f (x), contain the point x0 . ◼ ( ) is a maximum value of on I if for all x in I; x0 f f (x) ( ) ( ) 0 f x f x ◼ ( ) is a minimum value of on I if for all x in I; x0 f f (x) ( ) ( ) f x0 f x 2. Theorem If f (x) is continuous on a closed interval [a,b], then f (x) attains both a maximum and a minimum value there. If is continuous on a closed interval [a,b], then is f (x) bounded on [a,b]. f (x)
I.Max-Min TheoremNOTES2.Continuity1.Closedinterval-x+1,0≤x<111,x=1f(x)=^(0,a)f(x) =x-x+3,1<x≤202a01S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions 1. Closed interval (0, ) 1 ( ) a x f x = 2. Continuity I. Max-Min Theorem NOTES 0 a − + = − + = 3,1 2 1, 1 1, 0 1 ( ) x x x x x f x 0 1 2
Il.Zero Theorem1.DefinitionIf f(x,)= 0, then x,is calleda zero of f(x) .a0ba21S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions II. Zero Theorem 1.Definition If , then is called a zero of . ( ) 0 f (x) f x0 = x0 0 a b 0 b a 0 a b 0 b a x y 0 b a x y
Il. Zero Theorem2. TheoremIf f(x) is continuous on a closed interval [a,b], andf (a)f (b) < 0, then f(x)has at least one zero on (a,b) (a,b), s.t. f() = 0Note:Sufficientbut not necessary-00bxbx1S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions II. Zero Theorem 2. Theorem If is continuous on a closed interval [a,b], and f (a) f (b) < 0, then f (x) has at least one zero on (a,b) . f (x) (a,b), s.t. f ( ) = 0 Note：Sufficient but not necessary 0 b a x y 0 b a x y
Ill.Intermediate valueTheoremIf f(x)is continuous on a closed interval [a,bl, let Wbe a number between f(a) and f(b), then there is atleast one number between a and b such that f()=W.W52bS53&SS1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions III. Intermediate value Theorem If is continuous on a closed interval [a,b], let W be a number between f (a) and f (b), then there is at least one number between a and b such that =W. f (x) f ( ) W2 W1 1 2 3 a b
Ill.Intermediate value TheoremIf f(x)is continuous on a closed interval [a,bl, let Wbe a number between f(a) and f(b), then there is atleast one number between a and b such that f()=W.ProofLet F(x)= f(x)-WF(x) is continuous on [a,b],F(a)= f(a)-W and F(b)= f(b)-Wsince F(a). F(b)<0,therefore, 3 e (a,b), suchthat F()= 0.That is, f()=W.S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions III. Intermediate value Theorem If is continuous on a closed interval [a,b], let W be a number between f (a) and f (b), then there is at least one number between a and b such that =W. f (x) f ( ) Proof Let F(x) = f (x) −W F(x) is continuous on [a,b], F(a) = f (a) −W and F(b) = f (b) −W since F(a)F(b) 0, therefore, (a,b), such that F( ) = 0. That is, f ( ) = W
Example11. Use the Zero Theorem to show that the equationx-cosx=0hasasolutionbetweenx=0and x=元/2SolutionLet f(x)= x-cosx,then f(0)=0-1=-1, f(元 /2)=元 /2-cos元 /2=元/2.Since f(x) is continuous on[0,π / 2],then 3=(0,元/2),s.t. f()=0y元元元f(~0.0783COS44bisectionXx元0元元元元~-0.6075COS284888S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions Example 1 - cos 0 0 / 2. 1 x x = has a solution between x = and x = .Use the Zero Theorem to show that the equation Solution Let f (x) = x - cos x, then f (0) = 0 −1 = −1, f ( / 2) = / 2- cos / 2 = / 2. Since f (x)is continuouson[0, / 2], then (0, / 2), s.t. f ( ) = 0. 4 2 o x y 0.0783 4 cos 4 ) 4 ( = − f 8 0.6075 8 cos 8 ) 8 ( = − − f bisection
Example2Use the lntermediate Value Theoremto show that ona circular wire ring there are alwaystwo points oppositeformeachotherwith the sametemperature.(rcos,rsin0)元+0(-r,0)(r,0)(rcos(元+),rsin(元+0)AnalysisLet f(0) = T(r cos,r sinO)- T(r cos(π +),r sin(π +)f(O) = T(r,0)-T(-r,0),f(元)=T(-r,0)-T(r,0)S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions Example 2 form each other with the same temperature. a circular wire ring there are always two points opposite Use the Intermediate Value Theorem to show that on + (r cos ,rsin ) (r,0) (−r,0) (r cos( + ),rsin( + )) • • • • Let f ( ) = T(r cos,rsin )−T(r cos( + ),rsin( + )) f (0) = T(r,0) −T(−r,0), f ( ) = T(−r,0) −T(r,0) Analysis
Example3Let f(x) becontinuouson[0,2al, f(O)= f(2a)± f(a).Provethat3 e[0,2al,suchthat f()= f(a + )S1.10 Properties of Continuous Functions
§1.10 Properties of Continuous Functions [0 2 ], ( ) ( ). ( ) [0,2 ], (0) (2 ) ( ) = + = , a f f a f x a f f a f a Provethat suchthat Let b e continuouso n . Example 3