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?
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?
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:/