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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
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5+0 Answers

 #1
avatar
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
avatar+75017 
+2

 

 

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

 

 

 

cool cool cool

CPhill  May 20, 2017
 #4
avatar
0

Thx, how do I find that value?

Guest May 20, 2017
 #3
avatar+89633 
+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
avatar+75017 
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???

 

 

cool cool cool

CPhill  May 20, 2017

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