Answer to Question #180117 in Linear Algebra for Oliver

Question

Solve the following question by using bisection method.

upto six iteration

X³ - X - 11 = 0

Answer 1

Set

$f(x) = x^3 -x-11$

Consider when x = 2, then f(2) = -5 < 0

Also, consider when x = 2.5, then f(2.5) = 2.125>0

Since we have that f(2).f(2.5) < 0. We can state that the solution lies in the interval [2,2.5]

Iteration 1:

$m = \dfrac{2+2.5}{2} = 2.25$

f(2.25) = -1.8594 < 0

So we replace 2 with 2.25

Iteration 2:

$m = \dfrac{2.25+2.5}{2} = 2.375$

f(2.375) = 0.0215>0

So we replace 2.5 with 2.375

Iteration 3:

$m = \dfrac{2.25+2.375}{2} = 2.3125$

f(2.3125) = -0.946 < 0

So, we replace 2.25 with 2.3125

Iteration 4:

$m = \dfrac{2.3125+2.375}{2} = 2.3438$

f(2.3438) = -0.4684 < 0

So, we replace 2.3125 with 2.3438

Iteration 5:

$m = \dfrac{2.3438+2.375}{2} = 2.3594$

f(2.3594) = -0.2251 <0

So, we replace 2.3438 with 2.3594

Iteration 6:

$m = \dfrac{2.3594+2.375}{2} = 2.3672$

f(2.3672) = -0.1023<0

So the solution of $x^3 -x-11 = 0$ is 2.3672 in the interval [2,2.5], using bisection method up to 6 iterations.