# Thread: Newtons Method for finding Optimum

1. ## Newtons Method for finding Optimum

How do you use Newtons method to solve for a functions optimum? I know that for finding the root you use:
X1 = X0 - f(X0)/f'(X0)
X2 = X1 - f(X1)/f'(X1)
X3 = X2 - f(X2)/f'(X2)
...etc
For example lets say the function is: f(x) = 4x^3-2x^2 -5
What do you do differently to solve for optimum? My guess it has something to do with second derivative, but I'm not really sure. Any help would be appreciated

Thanks,
Brandon

2. Hi there, yes using the 2nd derivative is relevant here. You can use the first derivate as the new $f(x)$ and apply the method for $12x^2-4x = 0$

Do you follow?

3. so instead of using:
X1 = X0 - f(X0)/f'(X0)
X2 = X1 - f(X1)/f'(X1)
X3 = X2 - f(X2)/f'(X2)
...etc
I should use:
X1 = X0 - f'(X0)/f''(X0)
X2 = X1 - f'(X1)/f''(X1)
X3 = X2 - f'(X2)/f''(X2)
...etc
?

4. Bump

5. Yes, pls don't bump.

6. How do you know if you are finding the MIN or the MAX? For example lets say you have f(x) = x^2, and want to find the optimum with an initial guess of 5. i would do the following:

X1=5-2(5)/2=0
X2=0-2(0)/2=0
X3=0
Obviously this has a MIN at (0,0), and by applying Newtons Meathod you discover their is an Optimum at x=0, but if you Had no clue what a function looked like how would you know if the x value you are finding is its MIN or its MAX? Or in another case the function may have a MIN and a MAX, how do you know which one you found?

thanks

7. You could use the second derivative test to see whether that point is a minimum or a maximum point.