Prove that as a function of x, y never decreases

[Poirot]

Homework Statement

y= A(x-sin(x)) with A as a constant.

Homework Equations

dy/dx = A(1-cos(x)) ??

The Attempt at a Solution

If I am thinking about this correctly, one can just differentiate the function as I have, and argue that when the gradient (dy/dx) is less than zero, the function is decreasing. So from this:

A(1-cos(x))<0
so 1-cos(x)<0

cos(x)>1
which never happens for any x, including negative x since cos is an even function.

Is this the right way of doing this? I'm very rusty on proofs etc.

Thanks in advance!

 

[mfb]

That works, and you can prove it with the intermediate value theorem.

 

[Poirot]

Thank you very much indeed!

 

[Ray Vickson]

Homework Statement

y= A(x-sin(x)) with A as a constant.

Homework Equations

dy/dx = A(1-cos(x)) ??

The Attempt at a Solution

If I am thinking about this correctly, one can just differentiate the function as I have, and argue that when the gradient (dy/dx) is less than zero, the function is decreasing. So from this:

A(1-cos(x))<0
so 1-cos(x)<0

cos(x)>1
which never happens for any x, including negative x since cos is an even function.

Is this the right way of doing this? I'm very rusty on proofs etc.

Thanks in advance!

Your conclusion is correct only if $A >0$, which you did not state.

 

[Poirot]

Yeah sorry A is greater than zero, It's a combination of physical constants found earlier in the question.