Answer to Question #204965 in Quantitative Methods for Richa bhadauria

Question

Perform three iteration of the bisection method compute the positive root of

the equation f(x) = x² – 5x + 2

Answer 1

$$f(x) = x^2 – 5x + 2$$

Initial values $a_0=0, b_0=1$

$$f(a_0)=f(0)=(0)^2-5(0)+2=2$$

$$f(b_0)=f(1)=(1)^2-5(1)+2=-2$$

$$f(a_0)f(b_0)=2(-2)=-4<0$$

$$x_n=\dfrac{a_n+b_n}{2}$$

$$f(x_n)\leq\varepsilon=>answer=x_n$$

$$a_{n+1}=x_n, b_{n+1}=b_n, f(a_n)f(x_n)\geq 0$$

$$a_{n+1}=a_n, b_{n+1}=x_n, f(b_n)f(x_n)\geq 0$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} n & x & f(x) \\ \hline 1 & 0.5 & -0.25 \\ 2 & 0.25 & 0.8125 \\ 3 & 0.375 & 0.265625 \\ \hdashline \end{array}$$

Initial values $a_0=4, b_0=5$

$$f(a_0)=f(4)=(4)^2-5(4)+2=-2$$

$$f(b_0)=f(5)=(5)^2-5(5)+2=2$$

$$f(a_0)f(b_0)=-2(2)=-4<0$$

$$x_n=\dfrac{a_n+b_n}{2}$$

$$f(x_n)\leq\varepsilon=>answer=x_n$$

$$a_{n+1}=x_n, b_{n+1}=b_n, f(a_n)f(x_n)\geq 0$$

$$a_{n+1}=a_n, b_{n+1}=x_n, f(b_n)f(x_n)\geq 0$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} n & x & f(x) \\ \hline 1 & 4.5 & -0.25 \\ 2 & 4.75 & 0.8125 \\ 3 & 4.625 & 0.265625 \\ \hdashline \end{array}$$