document.write( "Question 133510: 3) Study the method of linear interpolation presented in chapter 5. Given p(x) = 2x^3 – 3x^2 – 6x + 4, show that there is an irrational root in the interval (0,1) and use linear interpolation to find the root accurate to two decimal places. \n" ); document.write( "
Algebra.Com's Answer #97679 by jim_thompson5910(35256)\"\" \"About 
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First let's find the possible rational roots of \"f%28x%29=2x%5E3-3x%5E2-6x+%2B+4\"\r
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\n" ); document.write( "\n" ); document.write( "Any rational zero can be found through this equation\r
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\n" ); document.write( "\n" ); document.write( " where p and q are the factors of the last and first coefficients\r
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\n" ); document.write( "\n" ); document.write( "So let's list the factors of 4 (the last coefficient):\r
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\n" ); document.write( "\n" ); document.write( "Now let's list the factors of 2 (the first coefficient):\r
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\n" ); document.write( "\n" ); document.write( "Now let's divide each factor of the last coefficient by each factor of the first coefficient\r
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\n" ); document.write( "\n" ); document.write( "Now simplify\r
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\n" ); document.write( "\n" ); document.write( "These are all the distinct rational zeros of the function that could occur\r
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\n" ); document.write( "\n" ); document.write( "So the only possible rational roots in the interval (0,1) is \"1%2F2\". However, if we plug in \"f%281%2F2%29\", we get \r
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\n" ); document.write( "\n" ); document.write( "\"f%281%2F2%29=2%281%2F2%29%5E3-3%281%2F2%29%5E2-6%281%2F2%29+%2B+4\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%2F2%29=1%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "So we can see that \"1%2F2\" is not a zero of \"f%28x%29=2x%5E3-3x%5E2-6x+%2B+4\". So there are no rational roots in the interval (0,1)\r
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\n" ); document.write( "\n" ); document.write( "Now let's see if there is a zero in the interval (0,1)\r
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\n" ); document.write( "\n" ); document.write( "Let's evaluate f(0) (the left endpoint of the interval)\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=2x%5E3-3x%5E2-6x%2B4\" Start with the given function\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=2%280%29%5E3-3%280%29%5E2-6%280%29%2B4\" Plug in \"x=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=2%2A0-3%2A0%5E2-6%2A0%2B4\" Raise 0 to the 3rd power to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=0-3%2A0%5E2-6%2A0%2B4\" Multiply 2 and 0 to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=0-3%2A0-6%2A0%2B4\" Raise 0 to the 2nd power to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=0-0-6%2A0%2B4\" Multiply 3 and 0 to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=0-6%2A0%2B4\" Subtract 0 from 0 to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=0-0%2B4\" Multiply 6 and 0 to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=0%2B4\" Subtract 0 from 0 to get 0\r
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\n" ); document.write( "\n" ); document.write( "\"f%280%29=4\" Add 0 and 4 to get 4\r
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\n" ); document.write( "\n" ); document.write( "So when \"x=0\", we have \"y=4\" (notice how y is positive)\r
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\n" ); document.write( "\n" ); document.write( "Now let's evaluate f(1) (the right endpoint of the interval) \r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=2x%5E3-3x%5E2-6x%2B4\" Start with the given function\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=2%281%29%5E3-3%281%29%5E2-6%281%29%2B4\" Plug in \"x=1\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=2%2A1-3%2A1%5E2-6%2A1%2B4\" Raise 1 to the 3rd power to get 1\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=2-3%2A1%5E2-6%2A1%2B4\" Multiply 2 and 1 to get 2\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=2-3%2A1-6%2A1%2B4\" Raise 1 to the 2nd power to get 1\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=2-3-6%2A1%2B4\" Multiply 3 and 1 to get 3\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=-1-6%2A1%2B4\" Subtract 3 from 2 to get -1\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=-1-6%2B4\" Multiply 6 and 1 to get 6\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=-7%2B4\" Subtract 6 from -1 to get -7\r
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\n" ); document.write( "\n" ); document.write( "\"f%281%29=-3\" Add -7 and 4 to get -3\r
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\n" ); document.write( "\n" ); document.write( "So when \"x=1\", we have \"y=-3\" (notice how y is negative)\r
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\n" ); document.write( "\n" ); document.write( "Since the sign of y changes from positive to negative as x goes from 0 to 1, this means that there must be a zero in the interval (0,1)\r
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\n" ); document.write( "\n" ); document.write( "Conclusion:\r
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\n" ); document.write( "\n" ); document.write( "Since we know that there is a zero in the interval (0,1), but it is not a rational zero, this means that the zero must be irrational.\r
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\n" ); document.write( "\n" ); document.write( "So there is an irrational zero in the interval (0,1)\r
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\n" ); document.write( "\n" ); document.write( "Now let's use linear interpolation to find the equation of the line that goes through 2 points on the curve\r
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\n" ); document.write( "\n" ); document.write( "Since we previously found that \"f%280%29=4\" and \"f%281%29=-3\", we have the two points (0,4) and (1,-3) that lie on the line. \r
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\n" ); document.write( "\n" ); document.write( "Now let's find the equation of the line through (0,4) and (1,-3)\r
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\n" ); document.write( "\n" ); document.write( "First lets find the slope through the points (\"0\",\"4\") and (\"1\",\"-3\")\r
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\n" ); document.write( "\n" ); document.write( "\"m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29\" Start with the slope formula (note: is the first point (\"0\",\"4\") and is the second point (\"1\",\"-3\"))\r
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\n" ); document.write( "\n" ); document.write( "\"m=%28-3-4%29%2F%281-0%29\" Plug in \"y%5B2%5D=-3\",\"y%5B1%5D=4\",\"x%5B2%5D=1\",\"x%5B1%5D=0\" (these are the coordinates of given points)\r
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\n" ); document.write( "\n" ); document.write( "\"m=+-7%2F1\" Subtract the terms in the numerator \"-3-4\" to get \"-7\". Subtract the terms in the denominator \"1-0\" to get \"1\"\r
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\n" ); document.write( "\n" ); document.write( "\"m=-7\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "So the slope is\r
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\n" ); document.write( "\n" ); document.write( "Now let's use the point-slope formula to find the equation of the line:\r
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\n" ); document.write( "\n" ); document.write( "------Point-Slope Formula------
\n" ); document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\" where \"m\" is the slope, and is one of the given points\r
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\n" ); document.write( "\n" ); document.write( "So lets use the Point-Slope Formula to find the equation of the line\r
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\n" ); document.write( "\n" ); document.write( "\"y-4=%28-7%29%28x-0%29\" Plug in \"m=-7\", \"x%5B1%5D=0\", and \"y%5B1%5D=4\" (these values are given)\r
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\n" ); document.write( "\n" ); document.write( "\"y-4=-7x%2B%28-7%29%280%29\" Distribute \"-7\"\r
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\n" ); document.write( "\n" ); document.write( "\"y-4=-7x%2B0\" Multiply \"-7\" and \"0\" to get \"0%2F0\". Now reduce \"0%2F0\" to get \"0\"\r
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\n" ); document.write( "\n" ); document.write( "\"y=-7x%2B0%2B4\" Add \"4\" to both sides to isolate y\r
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\n" ); document.write( "\n" ); document.write( "\"y=-7x%2B4\" Combine like terms \"0\" and \"4\" to get \"4\" \r
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\n" ); document.write( "\n" ); document.write( "So we have the line \"y=-7x%2B4\" that goes through (0,4) and (1,-3) which lie on the curve. \r
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\n" ); document.write( "\n" ); document.write( "\"0=-7x%2B4\" Now plug in \"y=0\"\r
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\n" ); document.write( "\n" ); document.write( "\"-4=-7x\" Subtract 4 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"-4%2F%28-7%29=x\" Divide both sides by 7 to isolate x\r
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\n" ); document.write( "\n" ); document.write( "So our answer is approximately \"x=0.571428571428571\"\r
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\n" ); document.write( "\n" ); document.write( "So the root to two decimal places is \"x=0.57\" \r
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