Using the** method of bisection**, find the root of the equation x^{3} – 9x + 1 = 0 between x = 2 and x = 4. Solve it up to four approximation four decimal places.

**Numerical Analysis, Engineering Mathematics **

Given equation is y = f(x) = x^{3} – 9x + 1 and it is continuous in the interval 2 ≤ x ≤ 4. Now, first determine the value of a function at the points 2 and 4 i.e.

f(2) = 2^{3} – 9(2) + 1 = 8 – 18 + 1 = – 9

f(4) = 4^{3} – 9(4) + 1 = 64 – 36 + 1 = +29

f(2) f(4) < 0

Since f(2) is negative and f(4) is a positive value, therefore the root f(x) lies in between 2 and 4. Now go to the First Approximation.

**First Approximation value of x**

x1 = (2+4)/2 = 3

Again, check the value of f(x) at x = 3,

f(3) = 3^{3} – 9(3) + 1 = 27 – 27 + 1 = 1

f(2)f(3) < 0

Since f(3) is a positive value and f(2) is a negative value, therefore the root of f(x) lies in between 2 and 3. Now go to Second Approximation.

**Second Approximation value of x**

x2 = (2+3)/2 = 2.5

Again, check the value of f(x) at x = 2.5,

f(2.5) = 2.5^{3} – 9(2.5) + 1 = – 5.87

f(2.5)f(3) < 0

Since f (3) is a positive value and f (2.5) is a negative value, therefore the root of f(x) lies in between 2.5 and 3. Now go to Third Approximation.

**Third Approximation value of x**

x3 = (2.5+3)/2 = 2.75

Again, check the value of f(x) at x = 2.75,

f(2.75) = 2.75^{3} – 9(2.75) + 1 = – 2.96

f(2.75)f(3) < 0

Since f(3) is a positive value and f(2.75) is a negative value, therefore the root of f(x) lies in between 2.75 and 3 and go to Fourth Approximation.

**Fourth Approximation value of x**

x4 = (2.75+3)/2 = 2.88

Again, check the value of f(x) at x = 2.88,

f(2.88) = 2.88^{3} – 9(2.88) + 1 = – 1.03

f(2.88)f(3) < 0

Since f(3) is a positive value and f(2.88) is a negative value, therefore the root of f(x) lies in between 2.88 and 3. Now go to Fifth Approximation.

**Fifth Approximation value of x**

x5 = (2.88+3)/2 = 2.94

Again, check the value of f(x) at x = 2.94,

f(2.94) = 2.94^{3} – 9(2.94) + 1 = – 0.05

f(2.94)f(3) < 0

Proceeding similarly we obtain x6 = 2.9375. Therefore, x = 2.9375 is our approximate root of f(x).

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