Question

Use the bissection method to approximate the root of f(x)=2x²-1 in the interval . Let ε=0.1 be the margin of error of approximation.

Answer 1

At first, we need to find the interval, where the roots are situated. $f(x)$ is a quadratic function. $f(0)=-1$ and $f(1)=f(-1)=1$. Since the function is continuous, it intersects $x$ -axis twice on the interval $[-1,1]$. Consider the interval $[0,1]$. The function intersects this interval once. I.e., the equation $f(x)=0$ has one root. We denote this root by $x_1$. Then, we set $x_2=-x_1$. It will be the second root. It follows from the fact that $f(x)=f(-x).$

We start the bisection method:

1. Divide the interval $[0,1]$ into two parts: $[0,0.5]$ and $[0.5,1]$. Consider $f(0.5)=-0.5.$ From the value of $f(0.5)$ we conclude that the root is in the interval $[0.5,1]$.
2. Divide the interval $[0.5,1]$ into two parts: $[0.5,0.75]$ and $[0.75,1]$. Consider $f(0.75)=0.125$. Thus, the root is in the interval $[0.5,0.75]$.
3. Divide the interval $[0.5,0.75]$ into two parts: $[0.5,0.625]$ and $[0.625,0.75]$. Consider $f(0.625)=-0.21875$. Thus, the root is in the interval $[0.625,0.75]$.
4. Divide the interval $[0.625,0.75]$ into two parts: $[0.625,0.6875]$ and $[0.6875,075]$. Consider $f(0.6875)=-0.0546875$. Thus, the root is in the interval $[0.6875,075]$. We can stop the method, because the length of the intervals is less than $0.1$. As an approximate value we set: $x_1=\frac{0.6875+0.75}{2}\approx0.7188$

Thus, the approximate value for one of roots is: $x_1\approx0.7188$. The approximate value for the second root is: $x_2\approx-0.7188$. We point out that exact values of roots are: $\frac{1}{\sqrt{2}}\approx0.7071$ and $-\frac{1}{\sqrt{2}}\approx-0.7071$.