As you may know, the overall purpose of Newton's method is to find the zero of a function, in other words the x-intercept. The process works like this:
1. Find the starting point of the Newton's method that is reasonably close to the zero.
ex) f(x) = x^3 - x^2 + 6x - 10
Reasonable first estimate: x = 2
2. Create a tangent line at that point.
ex) As you see in the image above, the tangent line at (2, 6) is y-6 = 14(x-2)
3. Solve for the point where the tangent line created at step 2 crosses the x-axis.
ex) The point the tangent line crosses the x-axis is (1.571,0)
4. Replace the starting point with the function value of the x-intercept of the created tangent line.
ex) Now, set this 1.571 as the new estimate and repeat step 1-4.
5. Repeat these processes until the points do not change much.
These images would be the later stages of repeated processes:
Hence, for this example, as the x value does not change between 1.487 and 1.488, we can conclude that the zero for this function would be between 1.487 and 1.488 (your calculator would be able to compute in more decimal places).
Hi, thank you for your question.
As you may know, the overall purpose of Newton's method is to find the zero of a function, in other words the x-intercept. The process works like this:
1. Find the starting point of the Newton's method that is reasonably close to the zero.
ex) f(x) = x^3 - x^2 + 6x - 10
Reasonable first estimate: x = 2
2. Create a tangent line at that point.
ex) As you see in the image above, the tangent line at (2, 6) is y-6 = 14(x-2)
3. Solve for the point where the tangent line created at step 2 crosses the x-axis.
ex) The point the tangent line crosses the x-axis is (1.571,0)
4. Replace the starting point with the function value of the x-intercept of the created tangent line.
ex) Now, set this 1.571 as the new estimate and repeat step 1-4.
5. Repeat these processes until the points do not change much.
These images would be the later stages of repeated processes:
Hence, for this example, as the x value does not change between 1.487 and 1.488, we can conclude that the zero for this function would be between 1.487 and 1.488 (your calculator would be able to compute in more decimal places).
Does this answer your question?