Consider the function f(x)=2x^3-3x^2-x+1 and | vedclass

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(C) To find the roots of $f(x)=2x^3-3x^2-x+1=0$,we check the sign of $f(x)$ at the endpoints of each interval using the Intermediate Value Theorem.$f(

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$f(x)=0$ has a root in every interval except in $I_4$

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(C) To find the roots of $f(x)=2x^3-3x^2-x+1=0$,we check the sign of $f(x)$ at the endpoints of each interval using the Intermediate Value Theorem.$f(-2) = 2(-8) - 3(4) - (-2) + 1 = -16 - 12 + 2 + 1 = -25$$f(-1) = 2(-1) - 3(1) - (-1) + 1 = -2 - 3 + 1 + 1 = -3$Since $f(-2)$ and $f(-1)$ have the same sign,there is no root in $I_4=[-2,-1]$.$f(0) = 1$Since $f(-1)=-3$ and $f(0)=1$,there is a root in $I_1=[-1,0]$.$f(1) = 2 - 3 - 1 + 1 = -1$Since $f(0)=1$ and $f(1)=-1$,there is a root in $I_2=[0,1]$.$f(2) = 2(8) - 3(4) - 2 + 1 = 16 - 12 - 2 + 1 = 3$Since $f(1)=-1$ and $f(2)=3$,there is a root in $I_3=[1,2]$.Thus,$f(x)=0$ has a root in every interval except $I_4$.

Consider the function $f(x)=2x^3-3x^2-x+1$ and the intervals $I_1=[-1,0]$,$I_2=[0,1]$,$I_3=[1,2]$,$I_4=[-2,-1]$. Then,

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