Increase and Decrease of a function question

[appplejack]

Homework Statement

f(x)=x — cosx, 0 ≤ x ≤ 2∏

Homework Equations

The Attempt at a Solution

I took the derivative of the function: f'(x)=1+ sinx, 0 ≤ x ≤ 2∏
I don't know how to determine whether it is increasing or decreasing between the interval given. What should I do?

 

[SammyS]

Homework Statement

f(x)=x — cosx, 0 ≤ x ≤ 2∏

Homework Equations

The Attempt at a Solution

I took the derivative of the function: f'(x)=1+ sinx, 0 ≤ x ≤ 2∏
I don't know how to determine whether it is increasing or decreasing between the interval given. What should I do?

What do you know about the range of sin(x) ?

 

[appplejack]

The Range of sin(x) is -1≤ x ≤ 1?

I have an additional question:

I saw this in my textbook and don't understand why this is true.

f (x) = -2sin2x — 2sinx = -2sinx(2cosx+1) <— how is this true?
Thanks

 

[SammyS]

The Range of sin(x) is -1≤ x ≤ 1?

I have an additional question:

I saw this in my textbook and don't understand why this is true.

f (x) = -2sin2x — 2sinx = -2sinx(2cosx+1) <— how is this true?
Thanks

So... Have you solve the original problem?

As for

f (x) = -2sin2x — 2sinx

= -2sinx(2cosx+1)​

The double angle identity for sine is: sin(2x) =2sin(x)cos(x)

Therefore: -2(2sin(x)cos(x)+sin(x)) =  ? 

 

[appplejack]

No. I drew the graph and it doesn't always increase but the answer says it increases from [0,2∏] How does knowing the range of sine solve help?

Actually, I've never learned "The double angle identity". I guess I have to know this then.

 

[SammyS]

Well, you got the derivative: f'(x)=1+ sin(x).

-1 ≤ sin(x) ≤ 1

Therefore:

0 ≤ 1+ sin(x) ≤ 2

So f' is positive except where sin(x) = -1, correct ?

Take it from there.

(It's a little tricky to figure out the behavior where f' is zero.)

 

[appplejack]

Thank you SammyS.