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...

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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.