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:root estimate = 1.75error bound = 0.25Step-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 < 0f(2) = 4 - 3.1 = 0.9 > 0Then there exists c in [0, 2], for which f(c) = 0 In the bisection method you generate a sequence [tex]x_n[/tex] 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 [tex]x_n = \frac{a+b}{2}[/tex]In this case: [tex]x_0 = 1[/tex][tex]f(0)<0\\f(1)<0\\f(2)>0[/tex]Then: [tex]x_1=1.5\\f(1.5)<0\\f(2)>0\\[/tex][tex]x_2= \frac{(1.5 + 2)}{2}= 1.75[/tex]The error bound is half the width of the interval [1.5, 2][tex]|error| < 0.25[/tex]