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Use Newton's method to find the absolute maximum value of the function f(x) = 2x sin x, 0 ≤ x ≤ π correct to six decimal places.



 Nov 6, 2015
edited by Guest  Nov 6, 2015

Best Answer 


Here's how to do it:


Newton's method

 Nov 6, 2015
Best Answer

Here's how to do it:


Newton's method

Alan Nov 6, 2015


I know this question is old but I wanted to look at it and this is the first chance I have had.

Alan has given a great answer but I wanted to do it without using a computer graphing program to help me.

(I did cheat and use some of Alan's results but that was just to save the necessity of number crunching)



\(f(x)=2xsinx   \qquad   0\le x\le \pi\\ f'(x)=2(sinx+xcosx)\\ f''(x)=2(cosx+cosx-xsinx) = 4cosx-2xsinx\\ \mbox{Now turning points will occur when f'(x)=0}\\ 2(sinx+xcosx)=0\\ sinx+xcosx=0\\ xcosx=-sinx\\ x=-tanx\\ \mbox{Now if I plot y=x and y=tanx then the point of intersection will be where this is true.}\\ \mbox{I can do a rough graph of this without the help of any graphing calculators..}\\ \mbox{I can see that the point of intersection is between } x=\pi/2 \;\;and x=\pi\\ \mbox{So like Alan said x=2 is a good first estimate.} \)



Now you use Newton's method to get closer and closer approximations.

I can never remember the formula so I have to derive it each time but with a little practice this is easy to do.


anyway, it is

\(\boxed{ x_1=x_0-\frac{f(x_0)}{f'(x_0)} }\\ Where \;\;x_0 \;\;\mbox{is the first estimate and }x_1 \mbox{ is the next estimate}\)


This is tedious but it can certainly be done by hand.   (you only need to do it 3 times.)

Anyway, Alan got the answer x=2.0288158.


Now you do need to ascertain what kind of a stationary point this is 

so you need to plug this valu of x into the second derivative.

It will give you   f''(2.0288158) < 0 

I have not done it because I cannot be bothered but I know this is true because Alan has kinly plotted f(x)for me and I can see that the concavity of the curve at the stat point is negative.


There you go.  That is how you do it without the need of graph aids.


I have graphed it just to help you understand what I was tallking about.  

I did graph it too but only roughly with a pen and paper.





 Nov 11, 2015

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