Felix's question at Yahoo Answers regarding Newton's method

[MarkFL]

Here is the question:

Suppose the line y = 5x − 2 is tangent to the curve y = f(x) when x = 4. If Newton's method is used to loc?

Suppose the line y = 5x − 2 is tangent to the curve y = f(x) when x = 4. If Newton's method is used to locate a root of the equation f(x) = 0 and the initial approximation is x1 = 4, find the second approximation x2.

x2 = ___?______4.8, 4

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Felix

Since the line $y=5x-2$ is tangent to the function $f(x)$ at $x=4$, we know two things:

a) The function and the line have a common point at $x=4$:

\(\displaystyle f(4)=5(4)-2=18\)

b) At $x=4$ the function's derivative is equal to the slope of the line:

\(\displaystyle f'(4)=5\)

Now, Newton's method gives us:

\(\displaystyle x_{n+1}=x_{n}-\frac{f\left(x_n \right)}{f'\left(x_n \right)}\)

If the initial approximation is \(\displaystyle x_1=4\), then the second approximation is:

\(\displaystyle x_{2}=x_{1}-\frac{f\left(x_1 \right)}{f'\left(x_1 \right)}=4-\frac{18}{5}=\frac{2}{5}\)

As we should expect, the second guess is simply the root of the tangent line:

\(\displaystyle 0=5x_2-2\implies x_2=\frac{2}{5}\)