a problem that uation log(x +1)+ log2 (x -6) -3. Write your solution in complet

Question

a problem that uation log(x +1)+ log2 (x -6) -3. Write your solution in complet öH le UI Ta nal Exams. Find all real solutions to the s a problem that appeared eq pon around). If there is any discrepancy between the solution set that a and the actual solution set of the given equ points of algebraic reasonung in your solution (you don't need to do txplaining the appears at the end of your process ation, try to give a logical explanation for the discrepancy The cubic equation 3 -9x70 has three real solutions: x-3.332, x 0.845, and x a 2487. Find all real solutions of each of the following equations. Since I'm only giving decimal approximations for the solutions to the original equation, turnabout is fair play; you don't need to come up with exact solu- tions. However, you should provide some algebraic justification for each solution; you're not allowed to just outsource the entire job to a graphing calculator 3. (a) (2y-5)-9(2y -5)+7 (b) sin3 t _ 9 sin t +7=0 0 Recall that two equations are equivalent if we can show that each equation

Explanation / Answer

Using newton method of root finding

x(n+1)=x(n)-f(x(n))/f'(x(n))

1)

f(x)=(2*x-5).^2-9*(2*x-5)+7;

f'(x)=4*(2*x-5)-18

using initial guess x=0;

Iteration f(x_n) diff(x_n) x_(n+1)  
1 77.000000 -38.000000 2.026316
2 16.423823 -21.789474 2.780066
3 2.272558 -15.759471 2.924269
4 0.083178 -14.605850 2.929964
5 0.000130 -14.560291 2.929973
6 0.000000 -14.560220 2.929973
7 0.000000 -14.560220 2.929973
8 0.000000 -14.560220 2.929973

thus one root is 2.929973

change the initial guess x=8;

Iteration f(x_n) diff(x_n) x_(n+1)  
1 29.000000 26.000000 6.884615
2 4.976331 17.076923 6.593209
3 0.339672 14.745669 6.570173
4 0.002123 14.561386 6.570027
5 0.000000 14.560220 6.570027
6 -0.000000 14.560220 6.570027
7 -0.000000 14.560220 6.570027
8 -0.000000 14.560220 6.570027

another root is 6.570027

2)

f(x)=sin3t -9sint+7

f'(x)=3sin2t*cos t -9cost

using initial guess t=0

Iteration f(t_n) diff(t_n) t_(n+1)  
1 7.000000 -9.000000 0.777778
2 1.030221 -5.359847 0.969989
3 0.137355 -3.933827 1.004905
4 0.004449 -3.679417 1.006114
5 0.000005 -3.670654 1.006116
6 0.000000 -3.670643 1.006116
7 0.000000 -3.670643 1.006116
8 0.000000 -3.670643 1.006116  

one root is 1.006116

using another guess t=3

Iteration f(t_n) diff(t_n) t_(n+1)  
1 5.732730 8.850786 2.352291
2 0.968950 5.274314 2.168580
3 0.125495 3.911698 2.136499
4 0.003754 3.678047 2.135478
5 0.000004 3.670651 2.135477
6 0.000000 3.670643 2.135477
7 -0.000000 3.670643 2.135477
8 -0.000000 3.670643 2.135477

root is 2.135477

using another guess t=6

Iteration f(t_n) diff(t_n) t_(n+1)  
1 9.492925 -8.416642 7.127875
2 0.688226 -4.861896 7.269431
3 0.074371 -3.815060 7.288925
4 0.001382 -3.673371 7.289301
5 0.000001 -3.670644 7.289301
6 0.000000 -3.670643 7.289301
7 0.000000 -3.670643 7.289301
8 0.000000 -3.670643 7.289301  

root is 7.289301

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