Newton's Method

[yojoe]

Consider f(x) = x^2sinx^2 on interval [0,3]

1)Use Newton's method to approx. to two decimal places the points x0 at which f achieves its global min.

2)Local max/min?

3)Inflection pts?


MY REASONING

1)I already found 1st and 2nd derivatives, but don't know how to proceed.

f'(x)=2x^3(cosx^2)+2xsinx^2
f''(x)=(-4x^4sinx^2)+(10x^2cosx^2)+2sinx^2

Instead of using: xn - (f(xn))/f'(xn)

would I use: xn - (f'(xn))/f''(xn) to help find global min?

2)Then for local max/min, I'm having trouble finding the roots from the 2nd derivative: setting it equal to zero
-4x^4sinx^2+10x^2cosx^2+2sinx^2

Am I on the right track?
Thanks in advance,
yojoe

 

[tkhunny]

The idea is to use the method to approximate the solutions, not to solve for the solutions. You must simply pick a starting value. As there may be no clue at all, simply pick anything in the Domain and see where it takes you. Something close to x = 0 or x = 1 often makes a good starting place.

You have the right idea: would I use: x_n+1 = x_n - (f'(x_n))/f''(x_n) to help find global min?

Newton's method doesn't care what function you have, so long as its first derivative exists. You are trying to find roots of the first derivative, so you have it correct. Pick a starting value and you're on your way.