Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 41

Given the function f(x)=cosx-x in the interval [0,4pi]
Now taking the derivative we get,
f'(x)=-sinx-1
To find the critical points, we have to equate f'(x)=0
therefore, -sinx-1=0
i.e sinx=-1
implies x=(3pi)/2, (7pi)/2 in the interval [0,4pi]
Now let us find the second derivative of f(x).
f''(x)=-cosx
Applying second derivative test we get,
f''(3 pi/2)=0
f''(7pi/2)=0
So here the second derivative test fails. So we have to apply the first derivative test.
i.e. f'(3pi/2)=-sin(3 pi/2)-1=0
f'(7pi/2)=-sin(7pi/2)-1=0
Therefore first derivative test also fails. So for f(x) we do not have a maxima or minima at the critical points.