+0

# ind the absolute maximum and absolute minimum of g(x)=sin(x)-x on [pi/2 , 2pi]

0
360
5

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
#2
+92368
+2

Note that   sin (3 pi/ 2)  = -1      and 3pi/2  lies in the requested interval

CPhill  May 20, 2017
#4
0

Thx, how do I find that value?

Guest May 20, 2017
#3
+94086
+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
#5
+92368
0

3 pi / 2      lies exactly   between  pi  and 2pi

pi  =  180°    and  2pi  =  360°    so    3pi/2  =  270°

Is this what you are asking???

CPhill  May 20, 2017