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1 CHAPTER 5 5.1 (a) A plot indicates that roots occur at about x = –1.6 and 5.6. 8 4 0 -5 0 5 10 -4 -8 (b) x= − 2.4 ± (2.4) 2 − 4(−0.6)(5.5) 2(−0.6) −1.62859 = 5.62859 (c) First iteration: 5 + 10 xr = = 7.5 2 5.62859 − 7.5 εt = × 100% = 33.25% 5.62859 εa = 10 − 5 × 100% = 33.33% 10 + 5 f (5) f (7.5) = 2.5(−10.25) = −25.625 Therefore, the bracket is xl = 5 and xu = 7.5. Second iteration: 5 + 7.5 xr = = 6.25 2 5.62859 − 6.25 7.5 − 5 εt = × 100% = 11.04% ε a = × 100% = 20.00% 7.5 + 5 5.62859 f (5) f (6.25) = 2.5(−2.9375) = −7.34375 Consequently, the new bracket is xl = 5 and xu = 6.25. Third iteration: 5 + 6.25 xr = = 5.625 2 5.62859 − 5.625 6.25 − 5 εt = × 100% = 0.06% ε a = × 100% = 11.11% 5.62859 6.25 + 5 5.2 (a) A plot indicates that a single real root occurs at about x = 0.45. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 Therefore, the new bracket is xl = 0.5 and xu = 0.75. The process can be repeated until the approximate error falls below 10%. As summarized below, this occurs after 4 iterations yielding a root estimate of 0.53125. iteration 1 2 3 4 xl 0.50000 0.50000 0.50000 0.50000 xu 1.00000 0.75000 0.62500 0.56250 xr 0.75000 0.62500 0.56250 0.53125 f(xl) -1.21875 -1.21875 -1.21875 -1.21875 f(xr) 2.83105 1.19498 0.10600 -0.52422 f(xl)×f(xr) -3.45035 -1.45638 -0.12918 0.63889 εa 33.33% 20.00% 11.11% 5.88% (c) False position: First iteration: xl = 0.5 f(xl) = –1.21875 xu = 1 f(xu) = 5 5(0.5 − 1) = 0.59799 xr = 1 − − 1.21875 − 5 f (0.5) f (0.59799) = −1.2175(0.75057) = −0.91475 Therefore, the bracket is xl = 0.5 and xu = 0.59799. Second iteration: xl = 0.5 f(xl) = –1.21875 xu = 0.59799 f(xu) = 0.75057 0.75057(0.5 − 0.59799) x r = 0.59799 − = 0.56064 − 1.21875 − 0.75057 0.56064 − 0.59799 × 100% = 6.661% εa = 0.56064 The process can be repeated until the approximate error falls below 0.2%. As summarized below, this occurs after 4 iterations yielding a root estimate of 0.55705. iteration 1 2 3 4 xl 0.5 0.5 0.5 0.5 xu 1.00000 0.59799 0.56064 0.55734 f(xl) -1.21875 -1.21875 -1.21875 -1.21875 f(xu) 5.00000 0.75057 0.07024 0.00610 xr 0.59799 0.56064 0.55734 0.55705 f(xr) 0.75057 0.07024 0.00610 0.00053 f(xl)×f(xr) -0.91475 -0.08561 -0.00744 -0.00064 εa 6.661% 0.593% 0.051% 5.4 (a) The graph indicates that roots are located at about –0.5, 2 and 4.7. 40 20 0 -2 0 2 4 6 -20 (b) Using bisection, the first iteration is PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 −1 + 0 = −0.5 2 f (−1) f (−0.5) = 29(2.125) = 61.625 xr = Therefore, the root is in the second interval and the lower guess is redefined as xl = –0.5. The second iteration is −0.5 + 0 = −0.25 2 − 0.25 − (−0.5) εa = 100% = 100% − 0.25 xr = f (−0.5) f (−0.25) = 2.125(−6.76563) = −14.37695 Consequently, the root is in the first interval and the upper guess is redefined as xu = –0.25. All the iterations are displayed in the following table: i 1 2 3 4 5 6 7 8 xl -1.0 -0.5 -0.5 -0.5 -0.50000 -0.46875 -0.45313 -0.45313 f(xl) 29.00000 2.12500 2.12500 2.12500 2.12500 0.85880 0.24273 0.24273 xu 0 0 -0.25 -0.375 -0.438 -0.43750 -0.43750 -0.44531 f(xu) -13.00000 -13.00000 -6.76563 -2.66992 -0.36206 -0.36206 -0.36206 -0.06107 xr -0.5 -0.25 -0.375 -0.43750 -0.46875 -0.45313 -0.44531 -0.44922 f(xr) 2.12500 -6.76563 -2.66992 -0.36206 0.85880 0.24273 -0.06107 0.09048 εa 100.00% 100.00% 33.33% 14.29% 6.67% 3.45% 1.75% 0.87% Thus, after eight iterations, we obtain a root estimate of −0.44922 with an approximate error of 0.87%, which is below the stopping criterion of 1%. (c) Using false position, the first iteration is xr = 0 − −13(−1 − 0) = −0.30952 29 − (−13) f (−1) f (−0.30952) = 29(−4.90027) = −142.10775 Therefore, the root is in the first interval and the upper guess is redefined as xu = –0.30952. The second iteration is x r = −0.30952 − εa = −4.90027(−1 − (−0.30952)) = −0.40933 29 − (−4.90027) − 0.40933 − (−0.30952) 100% = 24.383% − 0.40933 f (−1) f (−0.40933) = 29(−1.42411) = −41.29925 Consequently, the root is in the first interval and the upper guess is redefined as xu = –0.40933. All the iterations are displayed in the following table: i 1 2 xl -1 -1 f(xl) 29.00 29.00 xu 0 -0.30952 f(xu) -13 -4.90027 xr -0.30952 -0.40933 f(xr) -4.90027 -1.42411 εa 24.383% PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5 3 4 5 -1 -1 -1 29.00 29.00 29.00 -0.40933 -0.43698 -0.44430 -1.42411 -0.38199 -0.10024 -0.43698 -0.44430 -0.44621 -0.38199 -0.10024 -0.02615 6.327% 1.647% 0.429% Therefore, after five iterations we obtain a root estimate of –0.44621 with an approximate error of 0.429%, which is below the stopping criterion of 1%. 5.5 A graph indicates that a nontrivial root (i.e., nonzero) is located at about 0.93. 0.5 0 -0.5 0 0.5 1 1.5 -1 -1.5 Using bisection, the first iteration is 0.5 + 1 = 0.75 2 f (0.5) f (0.75) = 0.354426(0.2597638) = 0.092067 xr = Therefore, the root is in the second interval and the lower guess is redefined as xl = 0.75. The second iteration is 0.75 + 1 = 0.875 2 0.875 − 0.75 εa = 100% = 14.29% 0.875 xr = f (0.75) f (0.875) = 0.259764(0.0976216) = 0.025359 Because the product is positive, the root is in the second interval and the lower guess is redefined as xl = 0.875. All the iterations are displayed in the following table: i 1 2 3 4 5 xl 0.5 0.75 0.875 0.875 0.90625 f(xl) 0.354426 0.259764 0.097622 0.097622 0.042903 xu 1 1 1 0.9375 0.9375 f(xu) -0.158529 -0.158529 -0.158529 -0.0178935 -0.0178935 xr 0.75 0.875 0.9375 0.90625 0.921875 f(xr) 0.2597638 0.0976216 -0.0178935 0.0429034 0.0132774 εa 14.29% 6.67% 3.45% 1.69% Consequently, after five iterations we obtain a root estimate of 0.921875 with an approximate error of 1.69%, which is below the stopping criterion of 2%. The result can be checked by substituting it into the original equation to verify that it is close to zero. f ( x) = sin( x) − x 3 = sin(0.921875) − 0.921875 3 = 0.0132774 5.6 (a) A graph of the function indicates a positive real root at approximately x = 1.2. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8 1 2 3 1 1 1 3.00000 2.87500 2.79688 0.5 0.5 0.5 -0.03333 -0.02174 -0.01397 2.87500 2.79688 2.74805 -0.02174 -0.01397 -0.00888 -0.01087 -0.00698 -0.00444 2.793% 1.777% 7.81% 4.88% 3.05% Therefore, after three iterations we obtain a root estimate of 2.74805 with an approximate error of 1.777%. Note that the true error is greater than the approximate error. This is not good because it means that we could stop the computation based on the erroneous assumption that the true error is at least as good as the approximate error. This is due to the slow convergence that results from the function’s shape. 5.8 The square root of 18 can be set up as a roots problem by determining the positive root of the function f ( x) = x 2 − 18 = 0 Using false position, the first iteration is 7(4 − 5) = 4.22222 −2−7 f (4) f (4.22222) = −2(−0.17284) = 0.34568 xr = 5 − Therefore, the root is in the second interval and the lower guess is redefined as xl = 4.22222. The second iteration is 7(4.22222 − 5) = 4.24096 − 0.17284 − 7 4.24096 − 4.22222 εa = 100% = 0.442% 4.24096 x r = 4.22222 − Thus, the computation can be stopped after just two iterations because 0.442% < 0.5%. Note that the true value is 4.2426. The technique converges so quickly because the function is very close to being a straight line in the interval between the guesses as in the plot of the function shown below. 10 5 0 -5 4 4.5 5 5.9 A graph of the function indicates a positive real root at approximately x = 3.7. 10 0 0 1 2 3 4 5 -10 Using false position, the first iteration is PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 9 10.43182(0 − 5) = 1.62003 − 5 − 10.43182 f (0) f (1.62003) = −5(−4.22944) = 21.147 xr = 5 − Therefore, the root is in the second interval and the lower guess is redefined as xl = 1.62003. The remaining iterations are summarized below i 1 2 3 4 5 6 xl 0 1.62003 2.59507 3.34532 3.62722 3.71242 xu 5.00000 5.00000 5.00000 5.00000 5.00000 5.00000 f(xl) -5 -4.2294 -4.7298 -2.1422 -0.6903 -0.197 f(xu) 10.43182 10.43182 10.43182 10.43182 10.43182 10.43182 xr 1.62003 2.59507 3.34532 3.62722 3.71242 3.73628 f(xr) -4.22944 -4.72984 -2.14219 -0.69027 -0.19700 -0.05424 f(xl)×f(xr) 21.14722 20.00459 10.13219 1.47869 0.13598 0.01069 εa 37.573% 22.427% 7.772% 2.295% 0.639% The final result, xr = 3.73628, can be checked by substituting it into the original function to yield a near-zero result, f (3.73628) = (3.73628) 2 cos 3.73628 − 5 = −0.05424 5.10 Using false position, the first iteration is 7(4.5 − 6) = 5.01754 − 3.6875 − 7 f (4.5) f (5.01754) = −3.6875(−1.00147) = 3.69294 xr = 6 − Therefore, the root is in the second interval and the lower guess is redefined as xu = 5.01754. The true error can be computed as εt = 5.60979 − 5.01754 100% = 10.56% 5.60979 The second iteration is 7(5.01754 − 6) = 5.14051 − 1.00147 − 7 5.60979 − 5.14051 εt = 100% = 8.37% 5.60979 xr = 6 − εa = 5.14051 − 5.01754 100% = 2.392% 5.14051 f (5.01754) f (5.14051) = −1.00147(−1.06504) = 1.06661 Consequently, the root is in the second interval and the lower guess is redefined as xu = 5.14051. All the iterations are displayed in the following table: i 1 2 3 xl 4.50000 5.01754 5.14051 xu 6 6 6 f(xl) -3.68750 -1.00147 -1.06504 f(xu) 7.00000 7.00000 7.00000 xr 5.01754 5.14051 5.25401 f(xr) -1.00147 -1.06504 -1.12177 f(xl)×f(xr) 3.69294 1.06661 1.19473 εa, % εt, % 2.392 2.160 10.56 8.37 6.34 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10 4 5 6 7 5.25401 5.35705 5.44264 5.50617 6 6 6 6 -1.12177 -1.07496 -0.90055 -0.65919 7.00000 7.00000 7.00000 7.00000 5.35705 5.44264 5.50617 5.54867 -1.07496 -0.90055 -0.65919 -0.43252 1.20586 0.96805 0.59363 0.28511 1.923 1.573 1.154 0.766 4.51 2.98 1.85 1.09 Notice that the results have the undesirable feature that the true error is greater than the approximate error. This is not good because it means that we could stop the computation based on the erroneous assumption that the true error is at least as good as the approximate error. This is due to the slow convergence that results from the function’s shape as shown in the following plot (recall Fig. 5.14). 8 6 4 2 0 -2 4 4.5 5 5.5 6 -4 -6 5.11 (a) x r = (80)1 / 3.5 = 3.497357 (b) Here is a summary of the results obtained with false position to the point that the approximate error falls below the stopping criterion of 2.5%: i 1 2 3 4 xl 2 2.768318 3.176817 3.364213 xu 5 5 5 5 f(xl) -68.68629 -44.70153 -22.85572 -10.16193 f(xu) 199.5085 199.5085 199.5085 199.5085 xr 2.76832 3.17682 3.36421 3.44349 f(xr) -44.70153 -22.85572 -10.16193 -4.229976 f(xl)×f(xr) 3070.382 1021.686 232.258 42.985 εa 12.859% 5.570% 2.302% 5.12 A plot of the function indicates a maximum at about 0.9. 10 5 0 -0.5 -5 0 0.5 1 1.5 -10 In order to determine the maximum with a root location technique, we must first differentiate the function to yield f ' ( x) = −12 x 5 − 6.4 x 3 + 12 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 6.3 (a) Fixed point. The equation can be solved in two ways. The way that converges is x i +1 = 1.8 x i + 2.5 The resulting iterations are i εa xi 5 3.391165 2.933274 2.789246 2.742379 2.726955 2.721859 2.720174 2.719616 0 1 2 3 4 5 6 7 8 47.44% 15.61% 5.16% 1.71% 0.57% 0.19% 0.06% 0.02% (b) Newton-Raphson x i +1 = x i − i − x i2 + 1.8 x i + 2.5 − 2 x i + 1.8 xi 0 1 2 3 4 5 f (x ) 5 3.353659 2.801332 2.721108 2.719341 2.719341 -13.5 -2.71044 -0.30506 -0.00644 -3.1E-06 -7.4E-13 f'(x) εa -8.2 -4.90732 -3.80266 -3.64222 -3.63868 -3.63868 49.09% 19.72% 2.95% 0.06% 0.00% 6.4 (a) A graph of the function indicates that there are 3 real roots at approximately 0.2, 1.5 and 6.3. 20 10 0 -10 0 2 4 6 -20 (b) The Newton-Raphson method can be set up as x i +1 = x i − − 1 + 5.5 x i − 4 x i2 + 0.5 x i3 5.5 − 8 x i + 1.5 x i2 This formula can be solved iteratively to determine the three roots as summarized in the following tables: PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 i 0 1 2 3 4 i 0 1 2 3 4 i 0 1 2 3 4 xi 0 0.181818 0.213375 0.214332 0.214333 xi 2 1.555556 1.482723 1.479775 1.479769 xi 6 6.347826 6.306535 6.305898 6.305898 f (x ) -1 -0.12923 -0.0037 -3.4E-06 -2.8E-12 f'(x) εa 5.5 4.095041 100.000000% 3.861294 14.789338% 3.85425 0.446594% 3.854244 0.000408% εa f (x ) f'(x) -2 -0.24143 -0.00903 -1.5E-05 -4.6E-11 -4.5 -3.31481 28.571429% -3.06408 4.912085% -3.0536 0.199247% -3.05358 0.000342% f (x ) f'(x) -4 0.625955 0.009379 2.22E-06 1.42E-13 11.5 15.15974 14.7063 14.69934 14.69934 εa 5.479452% 0.654728% 0.010114% 0.000002% Therefore, the roots are 0.214333, 1.479769, and 6.305898. 6.5 (a) The Newton-Raphson method can be set up as x i +1 = x i − − 2 + 6 x i − 4 x i2 + 0.5 x i3 6 − 8 x i + 1.5 x i2 Using an initial guess of 4.2, this formula jumps around and eventually converges on the root at 0.474572 after 9 iterations: i 0 1 2 3 4 5 6 7 8 9 xi 4.2 -4.84912 -2.57392 -1.13753 -0.27274 0.20283 0.415354 0.470574 0.474552 0.474572 f (x ) f'(x) εa -10.316 -182.162 -52.4699 -14.737 -3.94412 -0.94341 -0.16212 -0.01021 -5.2E-05 -1.4E-09 -1.14 80.06397 36.52895 17.04115 8.293487 4.439071 2.935948 2.567567 2.541384 2.541249 186.61% 88.39% 126.27% 317.08% 234.47% 51.17% 11.73% 0.84% 0.00% The reason for the behavior is depicted in the following plot. As can be seen, the guess of x = 4.2 corresponds to a small negative slope of the function. Hence, the first iteration shoots to a negative value that is far from a root. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6 20 10 0 -10 0 2 4 6 8 -20 6.6 (a) A graph of the function indicates that the lowest real root is approximately −0.4: 60 40 20 -2 -20 0 2 4 6 (b) The stopping criterion corresponding to 3 significant figures can be determined with Eq. 3.7 is ε s = (0.5 × 102 − 3 )% = 0.05% Using initial guesses of xi–1 = –1 and xi = –0.6, the secant method can be iterated to this level as summarized in the following table: i xi−1 f(xi−1) 0 1 2 3 4 -1 -0.6 -0.46059 -0.41963 -0.41547 29.4 7.5984 1.725475 0.159224 0.003966 xi εa f(xi) -0.6 -0.46059 -0.41963 -0.41547 -0.41536 7.5984 1.72547499 0.15922371 0.00396616 9.539E-06 30.268% 9.761% 1.002% 0.026% 6.7 A graph of the function indicates that the first positive root occurs at about 1.9. However, the plot also indicates that there are many other positive roots. 2 1 0 -1 0 2 4 6 8 10 -2 -3 (a) For initial guesses of xi–1 = 1.0 and xi = 3.0, four iterations of the secant method yields i xi−1 0 1 2 3 1 3 -0.02321 -1.22635 f(xi−1) xi 3 -0.57468 -1.69795 -0.02321 -0.48336 -1.22635 -2.74475 0.233951 εa f(xi) -1.697951521 -0.483363437 13023.081% -2.744750012 98.107% -0.274717273 624.189% PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 7 4 0.233951 -0.27472 0.396366 -0.211940326 40.976% The result jumps to a negative value due to the poor choice of initial guesses as illustrated in the following plot: 1 0 -2 -1 0 2 4 6 -2 -3 Thereafter, it seems to be converging slowly towards the lowest positive root. However, if the iterations are continued, the technique again runs into trouble when the near-zero slope at 0.5 is approached. At that point, the solution shoots far from the lowest root with the result that it eventually converges to a root at 177.26! (b) For initial guesses of xi–1 = 1.0 and xi = 3.0, four iterations of the secant method yields i xi−1 0 1 2 3 4 1.5 2.5 2.356929 2.547287 2.526339 f(xi−1) -0.99663 0.166396 0.669842 -0.08283 0.031471 xi εa f(xi) 2.5 2.356929 2.547287 2.526339 2.532107 0.1663963 0.6698423 -0.0828279 0.0314711 0.0005701 6.070% 7.473% 0.829% 0.228% For these guesses, the result jumps to the vicinity of the second lowest root at 2.5 as illustrated in the following plot: 1 0 -2 -1 0 2 4 6 -2 -3 Thereafter, although the two guesses bracket the lowest root, the way that the secant method sequences the iteration results in the technique missing the lowest root. (c) For initial guesses of xi–1 = 1.5 and xi = 2.25, four iterations of the secant method yields i xi−1 0 1 2 3 4 1.5 2.25 1.927018 1.951479 1.944604 f(xi−1) -0.996635 0.753821 -0.061769 0.024147 -0.000014 xi 2.25 1.927018 1.951479 1.944604 1.944608 f(xi) 0.753821 -0.061769 0.024147 -0.000014 0.000000 εa 16.761% 1.253% 0.354% 0.000% For this case, the secant method converges rapidly on the lowest root at 1.9446 as illustrated in the following plot: PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8 1 0 -2 -1 0 2 4 6 -2 -3 6.8 The modified secant method locates the root to the desired accuracy after one iteration: x0 = 3.5 x0 + δx0 = 3.535 x1 = 3.5 − εa = f(x0) = 0.21178 f(x0 + δx0) = 3.054461 0.035(0.21178) = 3.497392 3.054461 − 0.21178 3.497392 − 3.5 × 100% = 0.075% 3.497392 Note that within 3 iterations, the root is determined to 7 significant digits as summarized below: i 0 1 2 3 x f (x ) 3.5 3.497392 3.497358 3.497357 0.21178 0.002822 3.5E-05 4.35E-07 dx x+dx 0.03500 0.03497 0.03497 0.03497 3.535 3.532366 3.532331 3.532331 f(x+dx) 3.05446 2.83810 2.83521 2.83518 f'(x) 81.2194 81.0683 81.0662 81.0662 εa 0.075% 0.001% 0.000% 6.9 (a) Graphical 8 4 0 -4 0 1 2 3 4 -8 Highest real root ≈ 3.3 (b) Newton-Raphson i 0 1 2 3 xi 3.5 3.365651 3.345112 3.344645 f (x ) 0.60625 0.071249 0.001549 7.92E-07 f'(x) 4.5125 3.468997 3.318537 3.315145 εa 3.992% 0.614% 0.014% PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.