Question

Find the interval of unit length that contains the smallest positive root of the
equation f(x) = x^3 −5x^2+1 = 0. Starting with this interval, find an interval of
length 0.05 or less that contains the root, by Bisection method.

Answer 1

Use Regular False method to determine where the smallest positive root is.

Choose two points a and b such that f(a) and f(b) are of opposite signs.

$$f(0)=1, \\ f(1)=-3.$$

With a=0 and b=1 find the (first) value of point of approximation:

$$x=x_1=\frac{af(b)-bf(a)}{f(b)-f(a)}=\frac{0\cdot(-3)-1\cdot1}{(-3)-1}=0.25.\\ \space\\ f(x_1)=0.703.$$

Therefore, the interval of unit length that contains the smallest positive root is [0;1].

Starting with this interval, find an interval of length 0.05 or less that contains the root, by Bisection method

Consider our interval $[x_L;x_R]=[0;1]$. Find the center of the interval and the value of the function at this point:

$$x_c=\frac{x_L+x_R}{2}=\frac{0+1}{2}=0.5,\\ \space\\ f(x_c)=-0.125.$$

Check at what part of the interval - $[x_L;x_C]$ or $[x_c;x_R]$ - there is a root: if the values of the function at the ends of one of these intervals are opposite, there is a root:

$$f(0)f(0.5)<0,\\f(0.5)f(1)>0.$$

Hence, we will only consider $[0;0.5]$. Perform the next iteration by bisection method:

$$x_c=\frac{0+0.5}{2}=0.25,\\ f(x_c)=0.703.\\ \text{Is the root on the left or right of } x_c?\\ f(0)f(0.25)>0,\\f(0.25)f(0.5)<0.$$

Hence, we will only consider [0.25;0.5]. Perform the next iteration:

$$x_c=\frac{0.25+0.5}{2}=0.375,\\ f(x_c)=0.35.\\ \text{Is the root on the left or right of } x_c?\\ f(0.25)f(0.375)>0,\\f(0.375)f(0.5)<0.$$

Hence, we will only consider [0.375;0.5]. Perform the next iteration:

$$x_c=\frac{0.375+0.5}{2}=0.438,\\ f(x_c)=0.127.\\ \text{Is the root on the left of } x_c\text{ or on the right?}\\ f(0.438)f(0.375)>0,\\f(0.438)f(0.5)<0.$$

Hence, we will only consider [0.438;0.5]. Perform the next iteration:

$$x_c=\frac{0.438+0.5}{2}=0.469,\\ f(x_c)=0.0034.\\ \text{Is the root on the left of } x_c\text{ or on the right?}\\ f(0.438)f(0.469)>0,\\f(0.469)f(0.5)<0.$$

But the length of both of these intervals is less than 0.05, therefore, the smallest positive root lies between 0.469 and 0.5.