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

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

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extreme f(x)=2sec(x)-tan(x)

extreme points

f (x) = 2 sec (x) - tan (x)

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{6} + 2 πn,

Increasing

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

Increasing

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

Decreasing

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

Decreasing

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

Plug the extreme point

x = \frac{π}{6} + 2 πn

into

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

Minimum (\frac{π}{6} + 2 πn, 3)

Plug the extreme point

x = \frac{5 π}{6} + 2 πn

into

2 sec (x) - tan (x) \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)

f (x) = \frac{x ^{2} - 9 x + 39}{x - 7}

4 x^{2} - 8 x + 16

x + \frac{3}{x}

f (x, y) = 6 x^{2} + y^{3} - 12 x y + 2

f (x) = x^{2} - 10

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