How Can We Determine the Max Min of Trig Functions in the Range (0, pi)?

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Discussion Overview

The discussion revolves around determining the maximum and minimum values of the function f(x) = sin(x) - x² within the interval (0, π). Participants explore various methods, including analytical approaches and numerical approximations, while addressing the behavior of the derivative f'(x) = cos(x) - 2x.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes setting the derivative f'(x) = 0 to find critical points, leading to the equation cos(x) - 2x = 0.
  • Another participant challenges the logic that cos(x) - 2x < 1 - 2x implies cos(x) - 2x = 0, stating that this is not a valid conclusion.
  • A request for advice on finding the maximum and minimum values of the function is made, indicating a need for further guidance.
  • One participant suggests using numerical approximation methods to solve the equation cos(x) - 2x = 0, providing a step-by-step approach for iteration.
  • Another participant inquires about bounding the derivative f'(x) on the interval (0, π) without explicitly finding the maximum and minimum values, providing some initial evaluations at the endpoints.
  • A further exploration of the behavior of sin(x) and its relationship to the function is presented, discussing its maximum slope and monotonic decrease.

Areas of Agreement / Disagreement

Participants express differing views on the validity of certain logical steps in determining critical points. There is no consensus on the best method to find the maximum and minimum values, with multiple approaches being discussed.

Contextual Notes

Some participants note the limitations of analytical solutions and the necessity of numerical methods, while others highlight the complexity of bounding the derivative without finding extrema.

Who May Find This Useful

Individuals interested in mathematical analysis, numerical methods, and the behavior of trigonometric functions within specified intervals may find this discussion relevant.

Somefantastik
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x in (0,pi];

f(x) = sin(x)-x2;

f'(x) = cos(x) - 2x;

f'(x) = 0 ==> cos(x) - 2x = 0;

since |cos(x)| ≤ 1,

cos(x) - 2x ≤ 1 - 2x;

Now 1-2x = 0 <==> x = 1/2;

f'(1/4) = cos(1/4) - 2*(1/4) > 0 and f'(3/4) = cos(3/4) - 2*(3/4) < 0;

==> x = 1/2 is maximum and f'(x) ≤ 1/2;

Is my logic correct?
 
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Somefantastik said:
x in (0,pi];

f(x) = sin(x)-x2;

f'(x) = cos(x) - 2x;

f'(x) = 0 ==> cos(x) - 2x = 0;

since |cos(x)| ≤ 1,

cos(x) - 2x ≤ 1 - 2x;

Now 1-2x = 0 <==> x = 1/2;
The fact that 1- 2x= 0 does NOT mean cos(x)- 2x= 0 since cos(x)- 2x< 1- 2x, if 1- 2x= 0 then cos(x)- 2x< 0. In fact, with x= 1/2, cos(1/2)- 2(1/2)= -1.22. cos(x)- 2x= 0 when cos(x)= 2x which happens when x is approximately .45.

f'(1/4) = cos(1/4) - 2*(1/4) > 0 and f'(3/4) = cos(3/4) - 2*(3/4) < 0;

==> x = 1/2 is maximum and f'(x) ≤ 1/2;

Is my logic correct?
No, it is not. saying that cos(x)- 2x< 1- 2x= 0 does not mean cos(x)- 2x= 0.
 
I'm trying to find the max and min of this function. Can someone give me some advice?
 
You'll have to do this by numerical approximation, since the equation is not solvable by analytic means.

cos (x) - 2x = 0 <==> x = .5 cos(x)

Start with an initial x value of your choice.
Evaluate .5*cos(x) at that value to get your next x value.
Repeat step 2.
When the input value and output value are "close enough" that's your approximate solution.
 
How about just getting a bound on this function ( f'(x) ) on (0,pi]; can I do that without finding the max and min?

f'(0) = 1 and f'(pi) = -1 - 2*pi

is my guess, but I am unsure of any bigger fluctuations on that interval.
 
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How about just getting a bound on this function ( f'(x) ) on (0,pi]; can I do that without finding the max and min?

Well let's look at the function shall we?

sinx is always less than 1.
the max slope of sinx occurs at zero and equals 1; thereafter it is always below the line y=x.
at small values

[tex]\sin x\quad \simeq x[/tex]

since [tex]x \le {(x)^2}[/tex]

and [tex]- {(x)^2}[/tex] decreases monotonically from 0 to [tex]- {(\pi )^2} \simeq 10[/tex]

The function is negative once [tex]x \le {(x)^2}[/tex] is greaer than 1.

Can you sketch it?
 
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