Question #306269

Find the real root of the equation 𝑥𝑒𝑥=2 correct upto two decimal places using the method of false position.


1
Expert's answer
2022-03-08T14:04:02-0500

f(x)=xex2f(x)=xe^x-2


For x1x-1 f(1)=1x12=0=0.718f(1)=1x^1-2=0=0.718


For x=0.5x=0.5 f(0.5)×e0.52=1.17f(0.5)×e^{0.5}-2=-1.17


As the value of f(x)f(x) at x=0.5x=0.5 is -ve and x=1.0x=1.0 it is positive. The root lies between x=0.5x=0.5 and x=1x=1


1st Step

x0=0.5x_0=0.5 and x1=1x_1=1


x2=x0(x1x0)f(x1)(x0)x_2=x_0-\frac{(x_1-x_0)}{f(x_1)-(x_0)} f(x0)f(x_0)


=0.510.50.718(1.17)=0.5-\frac{1-0.5}{0.718-(-1.17)} (1.17)(-1.17)


=0.8098=0.8098

f(x2)=f(0.8098)=0.8098e0.80982f(x_2)=f(0.8098)=0.8098e^{0.8098}-2

=0.8098=0.8098


2nd Step

Root lies between x1=1x_1=1 and x2=0.8098x_2=0.8098


x3=x2x1x2f(x1)f(x2)x_3=x_2-\frac{x_1-x_2}{f(x_1)-f(x_2)} f(x2)f(x_2)


=0.809810.80980.718+0.18=0 .8098-\frac{1-0.8098}{0.718+0.18} (0.18)(-0.18)

=0.8479=0.8479

f(x2)=f(0.8479)=0.8479e084792f(x_2)=f(0.8479)=0.8479-e^{0 8479}-2


=0.02=-0.02


3rd Step

Next root lies between x3=0.8479x_3=0 .8479 and x1=1x_1=1


x4=x3x1x3f(x1)f(x3)x_4=x_3-\frac{x_1-x_3}{f(x_1)-f(x_3)} f(x3)f(x_3)


=0.847910.84790.718(0.02)=0 .8479-\frac{1-0.8479}{0.718-(-0.02)} (0.02)(-0.02)

=0.8519=0.8519


f(x4)=0.8519e0.85192=0.0031f(x_4)=0.8519e^{0.8519}-2=-0.0031


4th Step

Next value of xx lies in between x1=1x_1=1 and x4=0.8519x_4=0.8519


x5=x4x1x4f(x1)f(x4)x_5=x_4-\frac{x_1-x_4}{f(x_1)-f(x_4)} f(x4)f(x_4)


=0.851910.85190.718+0.0031=0.8519-\frac{1-0.8519}{0.718+0.0031} (0.0031)=0.8525(0.0031)=0.8525


Approximate root to 4d.p =0.8525=0.8525






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