Question

Estimate the next guess for the root of x3 - 27 = 0 by using a modified bisection method as explained below. The initial bracket of (1.8,5.1) is found as a valid bracket. In the case of a modified bisection method, the root estimated at the end of the first iteration would be the point where the straight line drawn from the function at x = 1.8 to the function at x = 5.1 crosses the x-axis. What is this estimate of the root? (Keep 4 decimal places)

Answer 1

Bisection method for x=3 as the answer

Explanation:

Let's find an answer using the bisection method.

Because the problem gave us a valid bracket, we can establish that 'x', with a value between 1.8 and 5.1, is a solution for the equation, but let's understand why:

The equation
`x^3-27=0` can be written as:

`f(x)=x^3-27` then:

If
`x1=1.8` and
`f(x1)=(1.8)^3-27=-21.168`

If
`x2=5.1` and
`f(x2)=(5.1)^3-27=105.651`

Notice that the results are different in sign, the first one is negative whereas the second one is positive, meaning that there is a value (x) between them which permits f(x)=0, which in other words means that there is a root between 1.8 and 5.1.

The bisection method uses the mid-point (m) in each interaction between the two numbers defining the bracket (a,b), so:

`m=(b-a)/2` which in our case is:

`m=(5.1+1.8)/2=3.45` and

`f(m)=(3.45)^3-27=14.0636`

Notice that the result is positive but lesser than 105.651 from using x2, so now we change the bracket from (1.8,5.1) to (1.8,3.45) and we apply the method again:

`m=(b-a)/2` which in our case is:

`m=(3.45+1.8)/2=2.625` and

`f(m)=(2.625)^3-27=-8.9121`

Notice that now the result is negative but higher than -21.168 obtained from using x1, so now we change the bracket from (1.8,3.45) to (2.625,3.45). After this we apply the method again and again using the same estructure until we obtain a 'm' value from which f(m)=0, like this:

`m=(3.45+2.625)/2=3.0375` and
`f(m)=(3.0375)^3-27=1.0252`

`m=(2.625+3.0375)/2=2.83125` and
`f(m)=(2.83125)^3-27=-4.3048`

`m=(2.83125+3.0375)/2=2.934375` and
`f(m)=(2.934375)^3-27=-1.7334`

`m=(2.934375+3.0375)/2=2.9859` and
`f(m)=(2.9859)^3-27=-0.3779`

`m=(2.9859+3.0375)/2=3.0117` and
`f(m)=(3.0117)^3-27=0.3171`

`m=(2.9859+3.0117)/2=2.9988` and
`f(m)=(2.9988)^3-27=-0.0324`

`m=(2.9988+3.0117)/2=3.0053` and
`f(m)=(3.0053)^3-27=0.1420`

`m=(2.9988+3.0053)/2=3.00205` and
`f(m)=(3.00205)^3-27=0.05539`

`m=(2.9988+3.00205)/2=3.0004` and
`f(m)=(3.0004)^3-27=0.0115`

`m=(2.9988+3.0004)/2=2.9996` and
`f(m)=(2.9996)^3-27=-0.0108`

`m=(2.9996+3.0004)/2=3` and
`f(m)=(3)^3-27=0`

Finally look that f(m)=0, so the root is x=3.