Question

1. show/prove there is an interval of the form n,n + 1, where n is an integer, that contains a solution to the equation (x^{3}-4x + 2=0). explicitly state any theorem applied and show why you can apply the theorem. then use the bisection method to approximate the solution to (x^{3}-4x + 2 = 0) with an error less than (\frac{1}{4}).

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, \(\frac{1.625 + 1.75}{2}=1.6875\))