Given the function f(x) =2sec(`theta)`+tan(`theta)` in the interval 0<theta<2 pi

We have to find the critical points.

So let us first take the derivative of the function and evaluate it to zero.

`f'(theta)=2sec(theta)tan(theta)+sec^2(theta)=0`

`(2sin(theta))/(cos^2(theta))+1/(cos^2(theta))``=0`

`2sin(theta)+1=0`

`sin(theta)=-1/2```

implies, theta= 7pi/6 and 11pi/6 (that comes in the interval 0<theta<2 pi)

These are...

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Given the function f(x) =2sec(`theta)`+tan(`theta)` in the interval 0<theta<2 pi

We have to find the critical points.

So let us first take the derivative of the function and evaluate it to zero.

`f'(theta)=2sec(theta)tan(theta)+sec^2(theta)=0`

`(2sin(theta))/(cos^2(theta))+1/(cos^2(theta))``=0`

`2sin(theta)+1=0`

`sin(theta)=-1/2```

implies, theta= 7pi/6 and 11pi/6 (that comes in the interval 0<theta<2 pi)

These are the critical points.

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