## - Interval Bisection

With this method we repeatedly bisect the interval containing the root in order to improve the approximation.

LESSON 1

Given that \(x^3-3x^2=1-x\) has a root between 2 and 3, find this root to 1 decimal place using the bisection method.

SOLUTION

\[x^3-3x^2=1-x\]

Rewrite the equation in the form \(f(x)=0\)

\[x^3-3x^2+x-1=0\]

Let \(f(x)=x^3-3x^2+x-1\)

\[f(2)=2^3-3(2)^2+2-1=-3\]

\[f(3)=3^3-3(3)^2+3-1=2\]

- Determine the midpoint of this interval in which the root lies.

Mid – point of interval is 2.5

\[f(2.5)=2.5^3-3(2.5)^2+2.5-1=-1.625\]

- Since there is a sign change between 2.5 and 3 the root occurs within this interval. We therefore now bisect this interval. Mid – point is 2.75

\[f(2.75)=2.75^3-3(2.75)^2+2.75-1\]

\[ =-0.140625\]

- Due to sign changes the root must be in the interval 2.75 and 3.

Mid – point is 2.875

\[f(2.875)=2.875^3-3(2.875)^2+2.875-1=0.8418\]

- Due to sign changes the root is between 2.75 and 2.875

Mid – point is 2.8125

\[f(2.8125)=2.8125^3-3(2.8125)^2+2.8125-1=0.33\]

Therefore, root lies between 2.75 and 2.8125. When these 2 values are rounded to 1 decimal place the answers are both 2.8. Consequently, the root is approximately 2.8.