Question

Find the real root of the equation 3x = cos x+1 by bisection method

Answer 1

Answer:

Hence 0.607 is the required root of the Equation/function .

Step-by-step explanation:

let f(0) = -2 < 0

f(1) = 1.4596 > 0

f(0.5) = -0.377 < 0

f(0.6) = -0.025 < 0

f(0.7) = 0.335 > 0

Note :- This equation should be solved by scientific calculator .

- As a < 0 and b > 0

Now take a = 0.6 and b = 0.7

© 1 approximation :-

x = 0.6+0.7 / 2 -----------> x = 0.65 And ;

f(0.65) = 0.153 > 0

Now ;

a = 0.6 and b = 0.65

© 2 approximation :-

x = 0.6+0.65 / 2 -----------> x = 0.625 And ;

f(0.625) = 0.06403 > 0

Now ;

a = 0.6 and b = 0.625

© 3 approximation :-

x = 0.6+0.625 / 2 -----------> x = 0.6125 And ;

f(0.6125) = 0.019286 > 0

Now ;

a = 0.6 and b = 0.6125

© 4 approximation :-

x = 0.6+0.6125 / 2 -----------> x = 0.6O625 And ;

f(0.60625) = -0.00304050 > 0

Now ;

a = 0.60625 and b = 0.6125

© 5 approximation :-

x = 0.60625+0.6125 / 2 -----------> x = 0.6O9375 And ;

f(0.609375) = 0.0081191 > 0

Now ;

a = 0.60625 and b = 0.609375

© 6 approximation :-

x = 0.60625+0.609375 / 2 -----------> x = 0.6O78125 And ;

f(0.6078125) > 0

Now ;

a = 0.60625 and b = 0.6078125

© 7 approximation :-

x = 0.60625+0.6078125 / 2 -----------> x = 0.6O70312 And ;

f(0.6070312) < 0

Now ;

a = 0.60703125 and b = 0.6078125

- Hence 0.607 is the required root of the Equation by bisection method .

Thenks .