# Continuity ( monotony table )

mtayab1994

## Homework Statement

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

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

## 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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## Homework Statement

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

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

## 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.)

mtayab1994
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.

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Homework Helper
Gold Member
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 ?

Homework Helper
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.