`f(x) = sin(x) + cos(x), 0<=x<=2pi` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of...

`f(x) = sin(x) + cos(x), 0<=x<=2pi` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.

Asked on by enotes

Textbook Question

Chapter 4, 4.3 - Problem 13 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`f(x)=sinx+cosx`

differentiating,

`f'(x)=cosx-sinx`

Now let us find the critical points by setting f'(x)=0,

`cosx-sinx=0`

cosx=sinx

tanx=1

x=pi/4 , 5pi/4

Now let us check the sign of f'(x) to find the intervals of increase or decrease by plugging test value in the intervals (0,pi/4) , (pi/4,5pi/4) and (5pi/4,2pi)

`f'(pi/6)=cos(pi/6)-sin(pi/6)=(sqrt(3)-1)/2=0.36603`

`f'(pi)=cos(pi)-sin(pi)=-1-0=-1`

`f'(3pi/2)=cos(3pi/2)-sin(3pi/2)=0-(-1)=1`

Since f'(pi/6) is positive , so the function is increasing in the interval(0.pi/4)

f'(pi) is negative , so the function is decreasing in the interval (pi/4,5pi/4)

f'(3pi/2) is positive , so the function is increasing in the interval (5pi/4,2pi)

So the function has Local maximum at x=pi/4 and Local minimum at x=5pi/4

`f(pi/4)=sin(pi/4)+cos(pi/4)=1/sqrt(2)+1/sqrt(2)=sqrt(2)`

`f(5pi/4)=sin(5pi/4)+cos(5pi/4)=-1/sqrt(2)-1/sqrt(2)=-sqrt(2)`

Now to find the intervals of concavity and points of inflection, let us find the second derivative of the function,

`f''(x)=-sin(x)-cos(x)`

-sinx-cosx=0

tanx=-1

x=3pi/4 , 7pi/4 

Now let us find the sign of f''(x) by plugging test points in the intervals (0,3pi/4) , (3pi/4,7pi/4) and (7pi/4,2pi)

`f''(pi/2)=-sin(pi/2)-cos(pi/2)=-1-0=-1`

`f''(pi)=-sin(pi)-cos(pi)=-0-(-1)=1`

`f''(15pi/16)=-sin(15pi/16)-cos(15pi/16)=0.78569`

Since f''(pi/2) is negative , so the function is concave down in the interval (0,3pi/4)

f''(pi) is positive , so the function is concave up in the interval (3pi/4 , 7pi/4)

f''(15pi/16) is positive , so the function is concave up in the interval (7pi/4,2pi)

Since the concavity is changing so x=3pi/4 and 7pi/4 are the inflection points

We’ve answered 318,932 questions. We can answer yours, too.

Ask a question