As you have just seen with the function f (x) = cos x - x, Newton\'s method is f

Question

As you have just seen with the function f (x) = cos x - x, Newton's method is frequently an efficient way of finding a root to an equation. But Newton's method doesn't always work. Sometimes it gives you an answer you don't want, sometimes it can't be used, and sometimes it takes you further and further away from the root you are looking for! Consider f(x) = 3 + 2x - x^2. Use algebra to find the positive root of f. Now let's say that some people wanted to try Newton's method out on the above problem. What would happen if they started with x_1 = 0? What would happen if they started with x_1 = 1? Suppose f(x) was a very complicated function with a lot of roots in a very small interval. What problems could arise in trying to use Newton's method to find all the roots'? Now we consider an even worse scenario: one where Newton's method lakes you further away from the only root of a function, rather than closer. In order to understand the circumstances which cause this to happen, consider the following: Suppose you arc at a point x where the function is positive but the derivative is negative. Will the next iteration give you a number larger or smaller than x? Let f(x) = x/1 + x^2. What is the only root of f? Graph f on the interval [0, 10] using technology. Notice that it is first increasing, and then decreasing. What is the global maximum of f? What happens if we use Newton's method starting at x_1 = 2? Can you explain why Newton's method "thinks" it is approaching a root?

Explanation / Answer

1 f(x) = 3 +2x-x2=0 to find the roots we apply factorisation method

: 3 +3x-x-x2  3(1+x) - x(1+x) =0 => (3-x) (1+x) = 0 = > x=3 , -1

2 when we apply Newtons metod starting with x=0 we may get the -ve root x= -1

we may get the +ve root 3 when we start x= 1

3 we may not be able to identify the converging points when the roots are very close

5 f(x) = x/ 1+x2= 0 => x=0 is the only root

6 f '(x) = (1-x2) / [ (1+x2)2] =0 we get x= 1 , -1   

when xis in th interval (0,1) the value of f '(x) >0 the fn is increasing

in the interval ( 1,10) f '(x) <0 the fn is decreasing

hence x=1 is the point of local maximum

Local max of f (x) = f(1) = 1/2

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