Critical points for difficult function

Click For Summary

Discussion Overview

The discussion revolves around finding the number of relative maxima and minima of the function defined by its derivative f'(x) = [(8cos(x)/(x^2)-(⅛)] on the interval (1,10). Participants explore various methods to analyze the critical points, including analytical and graphical approaches.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the intermediate value theorem to argue that since f'(1) > f'(10), there must be at least one maximum or minimum in the interval (1,10).
  • Another participant questions the validity of the handwaving approach and proposes taking the complete derivative to find critical points.
  • A participant notes that setting the derivative equal to zero leads to the equation 64cos(x) - x^2 = 0, expressing uncertainty about how to solve it.
  • One participant proposes a factorization of the derivative, suggesting that it can be expressed as (8√(cos(x)) + x)(8√(cos(x)) - x) = 0, limiting the search for extrema to the interval [1, 5/2π].
  • Another participant expresses difficulty in solving the equation 8√(cos(x)) ± x = 0 and considers using a Maclaurin series for approximation.
  • A later reply indicates that exactness is not necessary if only the number of extrema is needed, suggesting that a simultaneous plot of cos(x) and x^2/64 could provide insight.
  • One participant expresses dissatisfaction with the inability to find an analytic solution, indicating a preference for a more rigorous approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a method to analytically solve for the critical points. There are multiple competing views on how to approach the problem, with some favoring graphical methods and others seeking an analytic solution.

Contextual Notes

Participants express uncertainty regarding the solvability of the derived equations and the implications of using approximations or graphical methods. The discussion highlights the limitations of the approaches considered and the varying levels of satisfaction with the methods proposed.

Robin64
Messages
34
Reaction score
3
Given that f'(x)=[(8cos(x)/(x2)-(⅛)], find the number of relative maxima and minima on the interval (1,10). Finding maxima and minima analytically wasn't fruitful for me, so instead I used a bit of handwaving. First I argued, using the intermediate value theorem, since f'(1)>f'(10), there exists at least one maximum or minimum. Then I said, well, the derivative is approximately equal to 8cos(x)/x2 and since cos(x) has 2 minima and one maximum on the interval (1,10), then f(x) must have 2 minima and one maximum on that same interval.

Is there way to solve this without that handwaving?
 
Physics news on Phys.org
Robin64 said:
Is there way to solve this without that handwaving?
Take the complete derivative and work it out ?
 
BvU said:
Take the complete derivative and work it out ?
The derivative is the expression in the problem. Setting the derivative equal to zero results in 64cos(x)-x2=0. I don't see the way forward to solving that expression.
 
$$ f' (x) = 0 \Leftrightarrow ( 8 \sqrt{\cos x} + x )( 8 \sqrt{\cos x} - x ) = 0 \quad ? $$
[edit] Limits the search for extrema to ## [1, {5\over 2} \pi] = [1, u]## with ##u < 8##
 
Last edited:
Yep, I got that. That's where I'm stuck, solving 8sqrt(cos(x)+/-x=0. I thought about expressing sqrt(cos(x)) as the first few terms of a Maclaurin series but that's less than exact or elegant.
 
Exactness isn't required if you only need the number of extrema. A simultaneous plot of ##\cos x## and ##x^2/64## does the trick...
 
I know. I was just unsatisfied with not being able to find an analytic solution.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
25
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K