SOLUTION: use the bisection method to find each of the zeros of the function: a) f(x)=x^2+x-1 on [-1, 1] b) f(x)= 2x^3-x-3 on [0,2] c) f(x)= sin x-x/2 on [1,3] d) f(x)=2^x-x-1 on [0,2]

Question 534295: use the bisection method to find each of the zeros of the function:
a) f(x)=x^2+x-1 on [-1, 1]
b) f(x)= 2x^3-x-3 on [0,2]
c) f(x)= sin x-x/2 on [1,3]
d) f(x)=2^x-x-1 on [0,2]
Answer by Edwin McCravy(20086) (Show Source): You can put this solution on YOUR website!

I'll just do c)

f(x)= sin(x)- on [1,3]

Method:
1. Make two columns with headings x and f(x)
2. evaluate f(x) for each of the two first values, they should
   have different signs.
3. To find the next value of x, average the last two values of x
   for which the values of f(x) had different signs:
4. Continue until three consecutive values for x when rounded to the
   desired number of decimal places are the same.

You didn't say how accurate, so I will get the zero to the nearest 
hundredths  

                                    x              f(x)
                                    1             .34147098
                                    3           -1.35888         
average 1&3                         2            -.0907026
average 1&2                         1.5           .24749499
average 2&1.5                       1.75          .10898595
average 2&1.75                      1.875         .01658578
average 2&1.875                     1.9375       -.0783219
average 1.875&1.9375                1.90625      -.0088639
average 1.875&1.90625               1.890625      .0039768
average 1.90625&1.890625            1.8984375    -.0024147
average 1.890625&1.8984375          1.890234375   .00429485
average 1.8984375&1.890234375       1.894335938   .00094806
average 1.8984375&1.894335938       1.896386719  -.0007313
average 1.894335938&1.896386719     1.895361329   .00010887
average 1.896386719&1.895361329     1.895874024  -.0003111

Those last three values of x rounded to the nearest hundredth
are both 1.90, so that is the value of the zero of f(x) 
between 1 and 3.

Edwin

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