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Are you sure that's the right equation? That has no roots in the interval [0,1].

well the original function was f(x)=2x(1-x^2+x)ln(x)=x^2-1, not sure if that changes things?

Ah. Then you should get \[2x(1-x^2+x)\ln(x)-x^2\large{\bf +1}.\] That might work better

yeah I made a typo in the original question... still I get stuck calculating x1 being equal to 0

Let me see what I get.

I get x1 = 0.5 - (-0.116434)/(-0.232868), which is 0.5 - 0.5 = 0

Hmm. I am getting the same thing as well. Perhaps try x=.49 or something.

it specifically says use x0=0.5. I just think my professor didn't double check his own work.

No problem.