## - 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.