Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 13

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