document.write( "Question 534295: use the bisection method to find each of the zeros of the function:
\n" ); document.write( "a) f(x)=x^2+x-1 on [-1, 1]
\n" ); document.write( "b) f(x)= 2x^3-x-3 on [0,2]
\n" ); document.write( "c) f(x)= sin x-x/2 on [1,3]
\n" ); document.write( "d) f(x)=2^x-x-1 on [0,2] \n" ); document.write( "

Algebra.Com's Answer #351459 by Edwin McCravy(20086) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "I'll just do c)\r\n" );
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document.write( "f(x)= sin(x)- on [1,3]\r\n" );
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document.write( "Method:\r\n" );
document.write( "1. Make two columns with headings x and f(x)\r\n" );
document.write( "2. evaluate f(x) for each of the two first values, they should\r\n" );
document.write( "   have different signs.\r\n" );
document.write( "3. To find the next value of x, average the last two values of x\r\n" );
document.write( "   for which the values of f(x) had different signs:\r\n" );
document.write( "4. Continue until three consecutive values for x when rounded to the\r\n" );
document.write( "   desired number of decimal places are the same.\r\n" );
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document.write( "You didn't say how accurate, so I will get the zero to the nearest \r\n" );
document.write( "hundredths  \r\n" );
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document.write( "                                    x              f(x)\r\n" );
document.write( "                                    1             .34147098\r\n" );
document.write( "                                    3           -1.35888         \r\n" );
document.write( "average 1&3                         2            -.0907026\r\n" );
document.write( "average 1&2                         1.5           .24749499\r\n" );
document.write( "average 2&1.5                       1.75          .10898595\r\n" );
document.write( "average 2&1.75                      1.875         .01658578\r\n" );
document.write( "average 2&1.875                     1.9375       -.0783219\r\n" );
document.write( "average 1.875&1.9375                1.90625      -.0088639\r\n" );
document.write( "average 1.875&1.90625               1.890625      .0039768\r\n" );
document.write( "average 1.90625&1.890625            1.8984375    -.0024147\r\n" );
document.write( "average 1.890625&1.8984375          1.890234375   .00429485\r\n" );
document.write( "average 1.8984375&1.890234375       1.894335938   .00094806\r\n" );
document.write( "average 1.8984375&1.894335938       1.896386719  -.0007313\r\n" );
document.write( "average 1.894335938&1.896386719     1.895361329   .00010887\r\n" );
document.write( "average 1.896386719&1.895361329     1.895874024  -.0003111\r\n" );
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document.write( "Those last three values of x rounded to the nearest hundredth\r\n" );
document.write( "are both 1.90, so that is the value of the zero of f(x) \r\n" );
document.write( "between 1 and 3.\r\n" );
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document.write( "Edwin

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