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We have the equation $f(x)=x^3-7x^2 + 14x -6 $ .I have to find the root of the equation using bisection method in the interval of $]1.3 , 2 [$

First I find $f(1.3)=2.567 >0 $ and $f(2)=2>0$

I guess this means that based on the Intermediate Value Theorem the function isn't continuous and there isnt a $p∈]1.3 , 2[$ such that $f(p)=0$?

Anyway $a0=1.3$ and $b0=2$ ..we have $p0=a0+ [(b0-a0)/2]=1.65$ ...And we have $f(po)=2.53$,which is positive,same as $f(a0)$ and $f(b0)$.

This means$ a1=po=1.65$,$b1=bo=2$...and I should go on with the iterations,but I dont understand,for how long should I go on?

user3543012
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  • Are you sure that a root exists in the given interval ? Remember that the function is continuous. – Claude Leibovici Dec 28 '14 at 15:19
  • Thats what I have written,since f(1.3)=2.567>0 and f(2)=2>0 ,this means that the function isnt continuous in the given interval,therefore we dont have a p∈ ]1.3 ,2[ such that f(p)=0 – user3543012 Dec 28 '14 at 15:26
  • The function is a polynomial so it is continuous, even more it's differentiable.. – Jacob Dec 28 '14 at 15:32
  • To answer partly the question for how long should I go on ?, I would answer don't start. You cannot start bisection if the interval does not contain the root. – Claude Leibovici Dec 28 '14 at 15:32
  • If the function does not change sign at the end points it does NOT mean that the function is not continuous. – Jacob Dec 28 '14 at 15:34
  • I understand.How do I prove that a root doesnt exist in the given interval? – user3543012 Dec 28 '14 at 15:36
  • In this case by luck we have the root $3$, so we can find nice expressions for all the roots. – André Nicolas Dec 28 '14 at 15:50

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