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

In summary, The problem involves finding a monotony table for a function g defined as g(x)=(1/4)x^2-sin(x) in the domain [-pi/2,pi/2]. However, upon calculating the first derivative, g'(x)=(1/2)x-cos(x), it was found that there is no exact solution for when g'(x) is equal to 0. This means that it is not possible to determine if the function increases or decreases in the given domain. However, by drawing a graph and observing the behavior of g'(x), an approximate boundary for monotony can be determined.
  • #1
mtayab1994
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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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  • #2
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.)
 
  • #3
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.
 
  • #4
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 ?
 
  • #5
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.
 
  • #6
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.
 

What is continuity?

Continuity is a concept in mathematics that refers to the smoothness and unbroken nature of a function or graph. It means that a small change in the input of a function will result in a small change in the output.

What is a monotony table?

A monotony table, also known as a monotonicity table, is a table that shows the change in a function's monotonicity (increasing or decreasing) over a given interval. It is used to analyze the behavior of a function and identify its critical points.

How is continuity determined?

Continuity is determined by three criteria: the function must be defined at the point in question, the limit of the function at that point must exist, and the limit must equal the function value at that point.

What is the difference between continuity and differentiability?

Continuity and differentiability are related but distinct concepts. Continuity refers to the smoothness and unbroken nature of a function, while differentiability refers to the ability to find the slope of the function at a given point. A function can be continuous but not differentiable at a certain point.

How are continuity and limits related?

Limits are used to determine continuity. If the limit of a function at a given point exists and equals the function value at that point, then the function is continuous at that point. In other words, continuity is a property that is defined using limits.

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