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