Thread: bisection root algorithm

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

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