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.
``
We’ll help your grades soar
Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.
- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support
Already a member? Log in here.
Are you a teacher? Sign up now