MHB 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 $x=4$:

$$f(4)=5(4)-2=18$$

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

$$f'(4)=5$$

Now, Newton's method gives us:

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

If the initial approximation is $$x_1=4$$, then the second approximation is:

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

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

$$0=5x_2-2\implies x_2=\frac{2}{5}$$