1. Newtons method question

Suppose the line y = 5x - 4 is tangent to the curve y = f(x) when x = 3. If Newton's method is used to locate a root of the equation f(x) = 0 and the initial approximation is x1 = 3, find the second approximation.

I am not sure how to find f(x). I tried finding the anti-derivative of the tangent line and used that as my f(x), but that didnt work.

2. Originally Posted by SyNtHeSiS
Suppose the line y = 5x - 4 is tangent to the curve y = f(x) when x = 3. If Newton's method is used to locate a root of the equation f(x) = 0 and the initial approximation is x1 = 3, find the second approximation.

I am not sure how to find f(x). I tried finding the anti-derivative of the tangent line and used that as my f(x), but that didnt work.
You don't need to know what $f(x)$ is! Recall that by Newton-Raphson Method, $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)}$.

You're given $x_1=3\implies x_2=x_1-\frac{f(x_1)}{f^{\prime}(x_1)}=3-\frac{f(3)}{f^{\prime}(3)}$. Note that $f(3)$ is the value of the tangent line at 3 (observe that the function and the tangent line have the same value at 3) and $f^{\prime}(3)$ is the slope of the given tangent line.

Can you finish the problem now?

3. So basically you indirectly using the value of f(x) and f'(x), by using the tangent equation, since it shares the same common point at x = 3?

4. Originally Posted by SyNtHeSiS
So basically you indirectly using the value of f(x) and f'(x), by using the tangent equation, since it shares the same common point at x = 3?
Exactly!!!

5. For a similar question (anti-derivatives) saying: Given the graph of f passes through the point (1, 6) and that the slope of its tangent line at (x, f(x)) is 2x + 1, find f(2).

In order to solve it, you had to find the anti-derivative of the tangent line. Howcome in the newtons method question this doesnt work, if it would work in this question?

6. These are completely different problems. Here you were specifically asked to find f(2). You have to find f(x), the anti-derivative, to do that.

With Newton's method you approximate a function by its tangent line and follow that tangent line to y= 0. At each step you don't want to find where f(x)= 0, just where the tangent line crosses y= 0.

7. Originally Posted by HallsofIvy
These are completely different problems. Here you were specifically asked to find f(2). You have to find f(x), the anti-derivative, to do that.

With Newton's method you approximate a function by its tangent line and follow that tangent line to y= 0. At each step you don't want to find where f(x)= 0, just where the tangent line crosses y= 0.
Oh ok, but howcome theres no subbing in of y = 0 for the tangent line in the newton method formula?