What is the extreme f(θ)=2sec(θ)-tan(θ) ?

The extreme f(θ)=2sec(θ)-tan(θ) is Minimum(pi/6+2pin,sqrt(3)),Maximum((5pi)/6+2pin,-sqrt(3))

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extreme f(θ)=2sec(θ)-tan(θ)

extreme

Minimum (\frac{π}{6} + 2 πn, 3), Maximum (\frac{5 π}{6} + 2 πn, - 3)

Solution steps

Find the critical points:

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Domain of

2 sec (θ) - tan (θ) : 2 πn \leq θ < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < θ < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < θ < 2 π + 2 πn

Find intervals:

Decreasing

: 2 πn \leq θ < \frac{π}{6} + 2 πn,

Increasing

: \frac{π}{6} + 2 πn < θ < \frac{π}{2} + 2 πn,

Increasing

: \frac{π}{2} + 2 πn < θ < \frac{5 π}{6} + 2 πn,

Decreasing

: \frac{5 π}{6} + 2 πn < θ < \frac{3 π}{2} + 2 πn,

Decreasing

: \frac{3 π}{2} + 2 πn < θ < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{6} + 2 πn

into

2 sec (θ) - tan (θ) \Rightarrow y = 3

Minimum (\frac{π}{6} + 2 πn, 3)

Plug the extreme point

x = \frac{5 π}{6} + 2 πn

into

2 sec (θ) - tan (θ) \Rightarrow y = - 3

Maximum (\frac{5 π}{6} + 2 πn, - 3)

Minimum (\frac{π}{6} + 2 πn, 3), Maximum (\frac{5 π}{6} + 2 πn, - 3)

What is the extreme f(θ)=2sec(θ)-tan(θ) ?
The extreme f(θ)=2sec(θ)-tan(θ) is Minimum(pi/6+2pin,sqrt(3)),Maximum((5pi)/6+2pin,-sqrt(3))