



 
Hi Tom, I would plot f(x) = x^{3}  270. Suppose x* is the point that satisfies f(x*) = 0. When I did this I found that x* is some point between x = 6.4 and x = 6.5. f(6.4) < 0 and f(6.5) > 0. In fact the graph shows me that x* is closer to 6.4 then 6.5. Now start your bisection search.
Continue for 8 iterations. For part (c) use 6.4 or 6.5 as the starting value and use the NewtonRaphson method to approximate x*. Harley  


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