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(a) How can i find the intervals on which f is increasing or decreasing?

(b) local max and min.

f(x)=sin(x)+cos(x), 0≤x≤2(pi)

 Dec 16, 2015

Best Answer 

 #1
avatar+130536 
+10

f(x) = sinx + cosx         take the derivative and set to 0

 

f ' (x)   = cosx  - sinx   = 0

 

So

 

cosx  = sin x        an this happens at    pi/4 and 5pi/4  on 0≤x≤2(pi)

 

Take the second derivative

 

f " (x)  = -sinx - cosx

 

Substituting in the critical points we have

 

-sin(pi/4) - cos(pi/4)   =   negative.....       so  we have a  relative max at pi/4

 

And

 

-sin (5pi/4)  - cos (5pi/4)   =   positive.......    so we have a relative minimum at 5pi/4

 

So the function is increasing on    [0, pi/4], decreasing on [pi/4, 5pi/4]  and increasing again  on [ 5pi/4, 2pi]

 

Here's a graph  [ in degrees]...........https://www.desmos.com/calculator/r1e6o1daoa

 

 

cool cool cool

 Dec 16, 2015
 #1
avatar+130536 
+10
Best Answer

f(x) = sinx + cosx         take the derivative and set to 0

 

f ' (x)   = cosx  - sinx   = 0

 

So

 

cosx  = sin x        an this happens at    pi/4 and 5pi/4  on 0≤x≤2(pi)

 

Take the second derivative

 

f " (x)  = -sinx - cosx

 

Substituting in the critical points we have

 

-sin(pi/4) - cos(pi/4)   =   negative.....       so  we have a  relative max at pi/4

 

And

 

-sin (5pi/4)  - cos (5pi/4)   =   positive.......    so we have a relative minimum at 5pi/4

 

So the function is increasing on    [0, pi/4], decreasing on [pi/4, 5pi/4]  and increasing again  on [ 5pi/4, 2pi]

 

Here's a graph  [ in degrees]...........https://www.desmos.com/calculator/r1e6o1daoa

 

 

cool cool cool

CPhill Dec 16, 2015

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