When feC¹ (R,R), and X*ER is a point with f(x*)=0 and f'(x*) ±0, then we know that there exists some 8>0 such that for every x₁=BS(x*), the parallel chord method converges to x*. What is the deeper reason why this is true? (You should know this reason, and you are simply asked to recognise it, so there is no need for proofs or counterexamples in this question.) a. For sufficiently small 8>0, the parallel chord method g(x)=x-f(x)/f'(x) is a contraction on B8(x*). b. For sufficiently small 8>0 the parallel chord method g(x)=x-f(x)/f'(xò) is a contraction from R into Bg(x*), i.e. for every XÃER, there exists N≥0 such that g(x)=Bg(x*) for all k≥N. C. I think that my friend thinks that the tutor thinks that the lecturer thinks that this statement is true. So my lecturer thinks that my tutor thinks that my friend thinks that I should tick this box. So this must be the box to tick. d. The parallel chord method g(x)=x-f(x)/f'(x) has the Newton property.

esc Today is Saturday. What day will it be 64 days from now? A. Saturday B. Monday OC. Sunday D. Friday Moving to another question will save this response. a L W M

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