Would anyone mind explaining part (a) of question 2F-1 from Problem Set 4? The answer key wasn't particularly helpful for me...
(I attached the original problem below in a reply)

Question

anonymous · Answer

attachment

phi · Answer

Here is a graph of the function and its derivative.

Here is the thinking
1)  when x is very negative e.g. -1000,  y= cos(x) - x implies  $ y \approx -x$ , and so y= + large number
2)  when x is very positive,  $ y \approx -x$, and so y= - large number
3) From 1 and 2 it follows y must be 0 at some intermediate point

4) the slope of the curve is *always* 0 or negative
   when the slope is 0, the curve moves "sideways"
   when the slope is negative, the curve moves "down"

5) From 4 we see the curve *never* goes up

6) conclusion,  the curve crosses the x-axis (from (3) ) ,  and does so only once (from (5) )

anonymous · Answer

This is how I approached the problem. (since it specifically asks to use the derivative to prove that there is only one root)

(1) y= cos (x) - x

(2) Therefore; y'= -sin (x) - 1

(3) In order for a graph to cross the x-axis more than once the graph must "turn around" at some critical point. (f' must change from positive to negative or negative to positive)

(4) So I found critical points where y' = 0. 
You end up with -1 = sin (x). 
This is true when x = 3pi/2 * n (n being any number of rotations around the unit circle)

(5) To check if statement (3) was true, I thought about [sin (x)] at [x= 3pi/2 + a very small amount] and [x = 3pi/2 - a very small amount] 
Essentially;  Lim (x -> 3pi/2) f' < 0 from both sides.

(6) This tells me that f' (the slope of the function) does not "turn around" at any point. 

(7) Going back to (3) we then know that the function can only cross the x-axis once since f(x) is decreasing for all x. 

To find the bounds, I zoomed in a bunch on my graphing calc and guessed.

I hope this helps!

phi · Answer

You did not say which part of 2F-1 you have a question about. For Newton's method, see http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx which might be more clear about this approach.