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

The extreme f(x)=2sec(x)+tan(x) is Maximum((7pi)/6+2pin,-sqrt(3)),Minimum((11pi)/6+2pin,sqrt(3))

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

extreme points

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

Maximum (\frac{7 π}{6} + 2 πn, - 3), Minimum (\frac{11 π}{6} + 2 πn, 3)

Solution steps

Find the critical points:

x = \frac{7 π}{6} + 2 πn, x = \frac{11 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:

Increasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Increasing

: \frac{π}{2} + 2 πn < x < \frac{7 π}{6} + 2 πn,

Decreasing

: \frac{7 π}{6} + 2 πn < x < \frac{3 π}{2} + 2 πn,

Decreasing

: \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn,

Increasing

: \frac{11 π}{6} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{7 π}{6} + 2 πn

into

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

Maximum (\frac{7 π}{6} + 2 πn, - 3)

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

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

Minimum (\frac{11 π}{6} + 2 πn, 3)

Maximum (\frac{7 π}{6} + 2 πn, - 3), Minimum (\frac{11 π}{6} + 2 πn, 3)

f (x) = 3 x - 6 cos (x), - 2 \leq x \leq 0

f (x) = \frac{3 x ^{2} - 8 x + 4}{x ^{2}}

f (x) = \frac{x ^{2} + 80}{x + 8}

f (x, y) = x^{2} + y^{2} - 2 x + 14 y - 11

f (x) = - 3 x^{2} + 30 x - 4

What is the extreme f(x)=2sec(x)+tan(x) ?
The extreme f(x)=2sec(x)+tan(x) is Maximum((7pi)/6+2pin,-sqrt(3)),Minimum((11pi)/6+2pin,sqrt(3))