Monotony Table for g in the Domain [-π/2,π/2]

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Homework Help Overview

The discussion revolves around the function g(x) = (1/4)x² - sin(x) and the task of creating a monotony table for this function within the domain [-π/2, π/2]. Participants are exploring the behavior of the function's derivative to determine intervals of increase and decrease.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants calculated the first derivative g'(x) = (1/2)x - cos(x) and attempted to solve for when it equals zero. There is a debate about the existence of solutions to this equation, with some asserting there are solutions while others express skepticism.

Discussion Status

The discussion is ongoing, with some participants suggesting the use of the Intermediate Value Theorem to argue for the existence of a solution in the interval (0, π/2). Others question the validity of this reasoning and seek clarification on the concept of a monotony table.

Contextual Notes

There is a mention of different methods to approach the problem, including graphical analysis, and a reference to external resources for understanding monotony tables. Participants are navigating through differing interpretations of the derivative's behavior.

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Homework Statement



Let g be a function defined as [tex]g(x)=(\frac{1}{4})x^{2}-sin(x)[/tex]

Give a monotony table for g in the domain [tex][-\frac{\pi}{2},\frac{\pi}{2}][/tex]



The Attempt at a Solution



I calculated the first derivative of g and i got g'(x)=(1/2)x-cos(x)

and then when I wanted to solve the equation (1/2)x-cos(x)=0 I found that there was no solution so we can't know if it increases or decreases.
 
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mtayab1994 said:

Homework Statement



Let g be a function defined as [tex]g(x)=(\frac{1}{4})x^{2}-sin(x)[/tex]

Give a monotony table for g in the domain [tex][-\frac{\pi}{2},\frac{\pi}{2}][/tex]



The Attempt at a Solution



I calculated the first derivative of g and i got g'(x)=(1/2)x-cos(x)

and then when I wanted to solve the equation (1/2)x-cos(x)=0 I found that there was no solution so we can't know if it increases or decreases.

There is a solution, because g'(0) = -cos(0) = -1, and g'(pi/2) = pi/4. As g' is continuous and has different signs at 0 and pi/2, it must be true that g'(x) = 0 for some x with 0 < x < pi/2. (Intermediate value theorem.)
 
jbunniii said:
There is a solution, because g'(0) = -cos(0) = -1, and g'(pi/2) = pi/4. As g' is continuous and has different signs at 0 and pi/2, it must be true that g'(x) = 0 for some x with 0 < x < pi/2. (Intermediate value theorem.)

I don't understand what you said because (1/2)x-cos(x)=0 has no solution.
 
mtayab1994 said:
I don't understand what you said because (1/2)x-cos(x)=0 has no solution.
Well, (1/2)x-cos(x)=0 has a solution at about x = 1.02987 .

What is a monotony table ?
 
SammyS said:
Well, (1/2)x-cos(x)=0 has a solution at about x = 1.02987 .

What is a monotony table ?


http://www.fmaths.com/studyoffunction/lesson.php

Take a look at this link it shows a monotony table.
 
mtayab1994 said:
and then when I wanted to solve the equation (1/2)x-cos(x)=0 I found that there was no solution so we can't know if it increases or decreases.

Hi mtayab1994!

There is a set of solutions.
Not an exact one, but an approximate one.

There are different ways to deal with problems like those.
The appropriate method here is probably by drawing a graph.

From the graph of g'(x), you can see where it increases monotonously, and where it decreases monotonously.
The exact boundary will be approximate.
 

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