`f(x) = cos(x) - x, [0, 4pi]` Find all relative extrema, use the second derivative test where applicable.

Textbook Question

Chapter 3, 3.4 - Problem 41 - Calculus of a Single Variable (10th Edition, Ron Larson).
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nees101's profile pic

nees101 | Student, Graduate | (Level 2) Adjunct Educator

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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.

balajia's profile pic

balajia | College Teacher | (Level 1) eNoter

Posted on

f(x)=cosx-x

f'(x) =-sinx-1

for extreme points f'(x)=0.

-sinx-1 = 0

sinx = -1

x = 3pi/2 , 7pi/2 . 

f''(x) = -cosx

f''(3pi/2)  =0

f''(7pi/2) = 0.

So f(x) has relative maxima at x = 3pi/2 and 7pi/2

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