Originally Posted by

**bobby77** find the root of f(x)=2x^3-x^2+x-1=0 to 3 decimal places using interval bisection.

First sketch the graph to locate the approximate position of the root.

See attachment:

Now we see the root is located at about . Now we need

a pair of values such that such that

.

It looks as though will do for these (check).

Now we need a table:

Code:

x_lo x_hi f(x_lo) f(x_hi) x_mid f(x_mid)
0.6 0.8 -0.328 0.184 0.7 -0.104

This is the first row, the next row is obtained by replacing or by , so that we

still bracket the zero, thus:

Code:

x_lo x_hi f(x_lo) f(x_hi) x_mid f(x_mid)
0.6 0.8 -0.328 0.184 0.7 -0.104
0.7 0.8 -0.104 0.184 0.75 0.0312
0.7 0.75 -0.104 0.0312 0.725 -0.0385

and so on utill .

RonL