explain why newtons method doesnt work for finding the solution of the equation ( x^{3}-3x + 8 = 0 ) if the…

explain why newtons method doesnt work for finding the solution of the equation ( x^{3}-3x + 8 = 0 ) if the initial approximation is chosen to be ( x_{1}=1 ). we have ( f(x)=x^{3}-3x + 8rightarrow f(x)=square ). if ( x_{1}=1 ), then ( f(x_{1})=square ) and the tangent line used for approximating ( x_{2} ) -select- attempting to find ( x_{2} ) results in -select- , which causes the method to fail.

explain why newtons method doesnt work for finding the solution of the equation ( x^{3}-3x + 8 = 0 ) if the initial approximation is chosen to be ( x_{1}=1 ). we have ( f(x)=x^{3}-3x + 8rightarrow f(x)=square ). if ( x_{1}=1 ), then ( f(x_{1})=square ) and the tangent line used for approximating ( x_{2} ) -select- attempting to find ( x_{2} ) results in -select- , which causes the method to fail.

Answer

Explanation:

Step1: Find the derivative of (f(x))

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (f(x)=x^{3}-3x + 8), we have (f^\prime(x)=3x^{2}-3).

Step2: Evaluate (f^\prime(x)) at (x = 1)

Substitute (x = 1) into (f^\prime(x)): (f^\prime(1)=3\times(1)^{2}-3=3 - 3=0).

Step3: Analyze the tangent - line

The formula for Newton's method is (x_{n+1}=x_{n}-\frac{f(x_{n})}{f^\prime(x_{n})}). When (f^\prime(x_{1}) = 0), the tangent line (y - f(x_{1})=f^\prime(x_{1})(x - x_{1})) becomes (y=f(1)) (since (f^\prime(x_{1}) = 0)), which is a horizontal line. Division by zero (in the formula (x_{2}=x_{1}-\frac{f(x_{1})}{f^\prime(x_{1})})) is undefined when trying to find (x_{2}).

Answer:

(f^\prime(x)=3x^{2}-3), (f^\prime(1) = 0), the tangent line is horizontal, and attempting to find (x_{2}) results in division by zero.