f'(x) = -sinx-1

f'(x)=0

-sinx-1=0

-sinx=1

sinx=-1

x=sin^-1 1

x= -pi/2

-pi/2 isnt in [pi/2 , 2pi]

what should i do to find the value?

Guest
May 20, 2017

#1
**0 **

I'm sorry.

**Find the absolute maximum and absolute minimum of f(x)=cos(x)-x on [pi/2 , 2pi]

f'(x) = -sinx-1

f'(x)=0

-sinx-1=0

-sinx=1

sinx=-1

x=sin^-1 1

x= -pi/2

-pi/2 isnt on [pi/2 , 2pi]

what should i do to find the value?

Guest
May 20, 2017

edited by
Guest
May 20, 2017

#3
**+1 **

f'(x) = -sinx-1

f'(x)=0

-sinx-1=0

-sinx=1

sinx=-1

x=sin^-1 1

x= -pi/2

-pi/2 isnt in [pi/2 , 2pi]

**Your question does not make sense to me**

**For instance:**

How can

f'(x) = -sinx-1

and

f'(x)=0

both at the same time?

that would only happen for x=-pi/2 (plus n revolutions where n is an integer)

Edit: I wrote this before you edited your question.

I do not know if it makes more sense now or not because I have not looked:/

Melody
May 20, 2017

edited by
Guest
May 20, 2017