# `f(theta) = 2sec(theta) + tan(theta), 0 < theta < 2pi` Find the critical numbers of the function. 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...

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.

``

Approved by eNotes Editorial Team