Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 35

To investigate what happens to maximum and minimum points we will take a look at the first derivative.
f'(x)=c+cos x
Maximum and minimum points are where f'(x)=0. Hence we get equation
c+cos x=0
cos x=-c
Let us now see where are extreme points for c=-1,0,1.
For c=-1
cos x=1=>x=2k pi, k in ZZ
For c=0
cos x=0=>x=pi/2+k pi, k in ZZ
For c=1
cos x=-1=>x=pi+2k pi, k in ZZ
From this we see that the first positive point of extreme moves from 0 to pi as c goes from -1 to 1.
However, when |c|>1 points of minimum and maximum vanish because equation cos x=-c has no solution (|cos x|leq 1, forall x in RR ).
To investigate what happens to inflection points we will take a look at second derivative.
f''(x)=-sin x
We can see that the second derivative does not contain c, therefore inflection points stay the same. Inflection points are where second derivative is equal to zero.
sin x=0
x=k pi, kin ZZ
The first image below shows the function for c=-1,0,1. Second image shows the function for c=1,2,5 . From those images we see that as c gets bigger linear part begins to dominate. For |c|=1 we get something like a sine wave under 45°. For c that is close to zero the graph looks similar to sine function. And when |c|> >1 (when c is significantly bigger than 1) graph looks similar to line. The bigger the c the closer the graph is to a line.