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Question Number 203668 by Perelman last updated on 25/Jan/24

Answered by witcher3 last updated on 25/Jan/24

x→^f (1/(4−3x)) f′(x)=(3/((4−3x)^2 )) ∀x∈[−1,(1/3)]⇒f(x)∈[(1/7),(1/3)].....(1) ⇒f[−1,(1/3)])⊂[(1/7),(1/3)] x_2 =(1/(4−3.2016))=(1/(−6044))∈[−1,(1/3)] ⇒∀n≥2 x_n ∈[−1,(1/3)] by (1) proof x_2 ∈[−1,(1/3)] supoose ∀n≥2 x_n ∈[−1,(1/3)] x_(n+1) =f(x_n )∈f[−1,(1/3)]⇒x_(n+1) ∈[(1/7),(1/3)]⇒x_(n+1) ∈[−1,(1/3)] ⇒x_n ∈[−1,(1/3)] x_(n+1) −x_n =(1/(4−3x_n ))−x_n =((3x_n ^2 −4x_n +1)/((4−3x_n )))=((3(x_n −(1/3))(x_n −1))/(4−3x_n )) x_n ∈[−1,(1/3)] ⇒x_(n+1) −x_n >0⇒x_n increaee Bounded sequence so cv since x→f(x) is continus x_(n+1) =f(x_n ) x_n cv to fix point of f lim_(n→∞) x_n =a⇒a=f(a)⇔a(4−3a)=1⇔a^2 −(4/3)a+(1/3)=0 ⇒a^2 −(1+(1/3))a+1.(1/3)=0 a∈{1,(1/3)} since a<(1/3)⇒lim_(n→∞) x_n =(1/3)

x→f14−3xf′(x)=3(4−3x)2∀x∈[−1,13]⇒f(x)∈[17,13].....(1)⇒f[−1,13])⊂[17,13]x2=14−3.2016=1−6044∈[−1,13]⇒∀n⩾2xn∈[−1,13]by(1)proofx2∈[−1,13]supoose∀n⩾2xn∈[−1,13]xn+1=f(xn)∈f[−1,13]⇒xn+1∈[17,13]⇒xn+1∈[−1,13]⇒xn∈[−1,13]xn+1−xn=14−3xn−xn=3xn2−4xn+1(4−3xn)=3(xn−13)(xn−1)4−3xnxn∈[−1,13]⇒xn+1−xn>0⇒xnincreaeeBoundedsequencesocvsincex→f(x)iscontinusxn+1=f(xn)xncvtofixpointoffDouble subscripts: use braces to clarify⇒a2−(1+13)a+1.13=0Double subscripts: use braces to clarify

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