Using Newton's Method to estimate a zero between a specific set of values? One of the problems in my Calculus books states, "Use Newton's method to estimate the zero between  $x=1$ and $x=2$ for the function $f(x)=X^3+2x-4$. Find the root to four decimal places."
Can someone help walk me through how to work this type of problem?
 A: The Newton Method recurrence for solving $f(x)=0$ is
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.\tag{1}$$ Here $f(x)=x^3+2x-4$ and $f'(x)=3x^2+2$.
Note that $f(1)=-1$ and $f(2)=8$. So it looks as if the root is closer to $1$ than to $2$. It is not unreasonable to make initial estimate $x_0=1$.
We now calculate $x_1$. By (1), we have
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=1+\frac{1}{5}.$$
Thus $x_1=1.2$. Repeat.
Run through a few iterations of the method. The only remaining issue is when to stop. Maybe when you do the calculations you will figure out when to stop. If you have difficulty, please report on your calculations and I will help you resolve this.
A: You atart with 2 as a rough initial approximation of a root. Now geometrically speaking the method asks you to draw a tangent to the curve $y=f(x)$, at the point with co-ordinates $\big(2,f(2)\big)$ and see where this tangent line cuts the $x$-axis, call that point $(x_1,0)$. This $x_1$ would be closer to the root than the initial value $2$. Now repeat the same procedure with $x_1$, that is draw the tangent to the same curve at $\big(x_1, f(x_1)\big)$ and so on.
The caclulation needs the equation of the tangent line: we know one point $(2,f(2))$. If the s0lpe of the tangent $m$ is known then the equation can be written as $y-f(2) = m(x-2)$. But calclusus says the the slope of the tangent line is the derivative: i.e., $m=f'(2)$. In this case as $f'(x)=3x^2+2$ we get $m=14$. So the tangent is the line given by $y-8=14(x-2)$.  Substitute $y=0$ in that equation. and get $x_1=\frac8{14}+2=  \frac{19}7$ as a better approximation. Starting with $2$ we have now arrived at $1.571428$.
Now keep repeating the procedure until the new number aggrees with the previous number in 4 decimal places.
