Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 44

S(x)=x-sin(x), 0leq x leq 4pi
(a)
Function is increasing (decreasing) when its derivative is greater (less) than zero. Therefore, we first need to find where the derivative is equal to zero.
S'(x)=1-cos(x)
1-cos(x)=0
cos(x)=1
x=2k pi, k in ZZ
Since 0 leq x leq 4pi and we are not interested in end points, we get
x=2pi
Now we know that the derivative has constant sign on these intervals
(0,2pi) and (2pi,4pi)
Now we simply choose one number from the first interval and check whether the derivative is greater or less than zero and than repeat the process for the second interval.
S'(\pi)=1-(-1)=2>0
S'(3pi)=1-(-1)=2>0
Therefore, the function is increasing for both of the intervals i.e. the function is increasing on [0,4pi].
(b)
Since the function is increasing over the whole domain minimum is achieved in left endpoint and maximum is achieved in right endpoint.
S(0)=0
Minimum is (0,0).
S(4pi)=4pi
Maximum is (4pi,4pi).
(c)
Function is convex (concave) when its second derivative is greater (less) than zero. So we will basically repeat the same process as in part a) but this time with second derivative.
S''(x)=sin(x)
sin(x)=0
x=k pi, k in ZZ
x=pi,\ 2pi,\ 3pi
This means that inflection points are at pi, 2pi, 3pi.
Therefore, our intervals are (0,pi), (pi,2pi), (2pi,3pi) and (3pi,4pi).
S(pi/2)=1>0
S((3pi)/2)=-1<0
S((5pi)/2)=1>0
S((7pi)/2)=-1<0

Function is convex on (0,pi) and (2pi,3pi), while concave on (pi,2pi) and (3pi,4pi).
Part (d) is in the image below.