Use Newton's method to find the absolute maximum value of the function f(x) = 6x sin x, 0 ≤ x ≤ π correct to six decimal places.

Accepted Solution

Determine the x-value at which the slope of the tangent line to the curve of f(x) = 6x*sin x is zero (horiz. tan. line).

f '(x) = 6[x*cos x+sin x * 1] = 0.  This, x*cos x + sin x = 0.  

Create the new function g(x) = x*cos x + sin x.  We want to find its roots/zeros.

Use the following formula (Newton's Method) to obtain roots/zeros:

x(n+1) = x(n) - -------                                                                                                                       f '(n)

Then for the given function's derivative, x*cos x + sin x,  we get

                         x*cos x + sin x                            x*cos x + sin x
x(n+1) = x(n) - -------------------------------- = x(n) - ------------------------
                         -x*sinx + cos x + cos x                -x*sinx + 2*cos x

Let the first (guessed) root be pi/6; this is between 0 and pi, as required.

                                                       (pi/6)*cos(pi/6) + sin(pi/6)
then the next root would be  pi/6 - ----------------------------------------
                                                        -(pi/y)*sin(pi/6) + 2*cos(pi/6)
                                 (pi/6)[sqrt(3)/2] + 1/2
which equals pi/6 - ----------------------------------------
                                  -(pi/6)*(1/2) + 2sqrt(3)/2

Lot of calculations!  Having set up this application of Newton's Method, I'm hoping you can continute this process and find one or more roots to 6 places.

Questions?  Please ask.