Question

2) (15 points) a) Determine the roots of f(x)=-12 – 21x +18r? - 2,75x' graphically. In addition, determine the first root of the function with b) bisection and c) false-position. For (b) and (c), use initial guesses of x, =-land x, = 0, and a stopping criterion of 1%. 3) (25 points) Determine the highest real root of f(x) = 2x – 11,7x² +17,7x-5 a) Graphically, b) Fixed-point iteration method (three iterations, x, = 3) c) Newton-Raphson method (three iterations, x, = 3) d) Secant method (three iterations, x_4 = 3 , Xo = 4) e) Modified secant method (three iterations, x, = 3, 8 = 0,01 ) Compute the approximate percent relative errors for your solutions.

Answer 1

40 20 (-0.04, 0) (16.485, 0) -40 -20 0 20 40 60 powe des

a) 612)=-12-21% +18X2-27543 graph of fin shown. 66-o.o 04)-0, ( 16.48 5) = 0 ho has two roots at ng= = 0.04 and at x2 = 16.485. Clo) Binection method: xe - 1, nuo O, E = 1% = 0.01 11+ iteration: f(-1) = 30270 610) =-12co By intermediate value theorem root lies between 1 1 .. and O. 20 = -1 to = -0.50 Хо -Ito fixo) = $1-0.5)- 37.3870 2nd iteration: 11-0.5) = 37-3870 and 120)=-12 Co :root lies between 0.5 and o 0-25 t1= -0-5 to 2 brand) = H(-0.25)= -1.33C0 Nowe, we make a table for all iterations. If(xe) If law) | c= xetxu /f(e) update hlal au O ae C b=C a-C 37-38 -1.33 12.91 4.71 1.45 a=C a-C с 6 -0.28 1 -1 302 -12 2 -0.5 37:38 0 -12 3 -0.5| 37:381-0.25 -1.33 4 -0.38 12.91 -0.25 -1.33 -0.31 4.71 -0.25 -1.33 1.45 -0.25 -1.33 7 -0.27 0 -0.25 -1.33 8 -0.27 0 -0.26 -0.68 0:27 0 6 -0.26 -0.17 11-0.27 -0.08 -0.27 o O -0.27 -0.0 13-0.27 0 -0.271 -0.02 CScanned with CamScanner D -0.5 -0.25 -0.38 -0-31 -0.28 -0.27 -0.26 -0.26 -0.26 -0.26 -0.27 -0.27 -0.27 b=c b=0 bec -0.26 -0.34 1-0 68 |-0.34 1-017 -0.000 1-0.04 -0.02 1o 0.27 b=0 bal b=0 b=c 121-0-27 b=0 b=0

approximate root as -0.27 (M)=0 so bisection metrod provides tre Equations fins Ami 0.27) q ist iteration! so x2 =112 -12-302 (C) False-ponition method'.. 'root lies between te 2-1 and Xuro xi-free). xu xe fraul-final -|- 302 x ot = -0.04 bin2 )= -11.1620 2nd: 61-1)=30270, H1-0.04) --11:16 LO ho x₂ = re- flee) xu-xe blow-boxel 7n3 = -1-302 X 0.04 +1 =-0.07 11:16 – 302 o 6143) = -10.28 CO 3rd: 61-1) = 302 30, 6(-0.07 ) =-10-2840 lux 4 = -| -- 302 -0.07 + 1 ~0.1 - 10.28-302 1194)= -9.3520 4 th: 61-11 = 302 >o & b1-0.1) - - 9.3560 0.1+1 15 :-)- 302 ( = -0.13 -9.35-302 Hlass) = -8.3620 5th: x6 =-1-302 -0.13+) -0.15 -8.36-302 bimc)= -7.3120 (.' bl-1)0 61-0.13)<0) ( ) option) CScanned with CamScanner

6th 61-1270 =-0.19 soro ) 61-1) >0 sbo.15)co hony = -1-302-60.15 +1) = -0.17 -736- 302 7th: 61-1870, 61-0.17) Co - 1 - 302 -0.17+) –6:38–302 B190) = 61-0.19) =-5.44 co 8 th: b(-1) 70, f(-0.19 ) co -0.19 +1 - 1 - 302 X g. -0.2 - 5.44-302 bing) = 61-0.2) = -4.5720 1-1) = 30 270 H1-0.2) --4.57 <0 ho xy = r--302 * to zgod) -0.22 119,0) = 6 -0.22 ) = -3.0C0 loth: 11-1)>0, 61-0.22)co so ^, = -1-302 ( - -0.22+1 -3.8 -302 gth. ) x 1 C-0.23 feli) f(-0.23) a-3-12 CO 61-1) 7o and f(-0.23) <0 23-|- 302 (-0.23 +1) -3-12-302 -0.23 so x12 2x12 Ami /root = -0.23) by method of fahre position. Scanned with Camscom
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