Question

Let f(x)=x^2- 3. 1. Show that [0, 2] is a bracketing interval for f 2. Carry out 3 iterations of the bisection method applied to f with initial bracketing interval 0, 2]. Give your estimate for the root and the error bound

Answer 1

Answer:

root estimate = 1.75

error bound = 0.25

Step-by-step explanation:

f is a polynomial, so it is continuous in R (real numbers). Then you can use Bolzano's theorem.

f(0) = -3.1 < 0

f(2) = 4 - 3.1 = 0.9 > 0

Then there exists c in [0, 2], for which f(c) = 0

In the bisection method you generate a sequence $x_n$ of approximations of a root. If you have a bracketing interval [a, b], such that

f(a) and f(b) have opposite signs, then you use  approximate the root as $x_n = \frac{a+b}{2}$

In this case:

$x_0 = 1$

$f(0)<0\\f(1)<0\\f(2)>0$

Then:

$x_1=1.5\\f(1.5)<0\\f(2)>0$

$x_2= \frac{(1.5 + 2)}{2}= 1.75$

The error bound is half the width of the interval [1.5, 2]

$|error| < 0.25$