What is the extreme 2sec(t)+tan(t) ?

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

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extreme 2sec(t)+tan(t)

extreme points

2 sec (t) + tan (t)

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: 2 πn \leq t < \frac{π}{2} + 2 πn,

Increasing

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

Decreasing

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

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{7 π}{6} + 2 πn

into

2 sec (t) + tan (t) \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 (t) + tan (t) \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) = \frac{e ^{3 x}}{3 x - 3}

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

f (x) = \frac{1}{4} x^{4} + \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 4

f (x) = x y^{2} = 9

f (x) = sin (θ)

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