3. show/prove there is an interval of the form n,n + 1, where n is an integer, that contains a solution to…

Explanation:

Step1: Check function values at endpoints

Let (f(x)=x^{3}-4x + 2). Evaluate (f(n)) and (f(n + 1)) for different integer values of (n). For (n = 1), (f(1)=1^{3}-4\times1 + 2=1 - 4+2=-1). For (n = 2), (f(2)=2^{3}-4\times2 + 2=8 - 8+2 = 2). Since (f(x)) is a polynomial, it is continuous everywhere (by the fact that polynomials are continuous functions - the Continuity of Polynomial Functions Theorem). And (f(1)\cdot f(2)=(-1)\times2=-2<0), so by the Intermediate - Value Theorem, which states that if (y = f(x)) is continuous on the closed interval ([a,b]) and (k) is a number between (f(a)) and (f(b)), then there exists at least one number (c) in the open interval ((a,b)) such that (f(c)=k). In our case, (k = 0), (a = 1), (b = 2), so there is a root of (f(x)=x^{3}-4x + 2) in the interval ([1,2]).

Step2: Apply the bisection method

The bisection method formula for the (n) - th approximation (x_n) of the root in the interval ([a_n,b_n]) is (x_n=\frac{a_n + b_n}{2}). Initial interval ([a_0,b_0]=[1,2]), (x_0=\frac{1 + 2}{2}=1.5), (f(1.5)=(1.5)^{3}-4\times1.5 + 2=3.375-6 + 2=-0.625). Since (f(1.5)\cdot f(2)<0), the new interval ([a_1,b_1]=[1.5,2]). (x_1=\frac{1.5 + 2}{2}=1.75), (f(1.75)=(1.75)^{3}-4\times1.75 + 2=5.359375-7 + 2=0.359375). Since (f(1.5)\cdot f(1.75)<0), the new interval ([a_2,b_2]=[1.5,1.75]). (x_2=\frac{1.5+1.75}{2}=1.625), (f(1.625)=(1.625)^{3}-4\times1.625 + 2=4.296875-6.5 + 2=-0.203125). Since (f(1.625)\cdot f(1.75)<0), the new interval ([a_3,b_3]=[1.625,1.75]). The length of the interval ([a_3,b_3]) is (b_3 - a_3=1.75-1.625 = 0.125=\frac{1}{8}<\frac{1}{4}).

Answer:

An approximation of the root is (x\approx1.6875) (the mid - point of the last interval ([1.625,1.75]) can be taken as an approximation, (\frac{1.625 + 1.75}{2}=1.6875))