What is the extreme 2sin(x-pi/6) ?

The extreme 2sin(x-pi/6) is Maximum((2pi)/3+2pin,2),Minimum((5pi)/3+2pin,-2)

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extreme 2sin(x-pi/6)

extreme points

2 sin (x - \frac{π}{6})

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

Solution steps

Find the critical points:

x = \frac{2 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Domain of

2 sin (x - \frac{π}{6}) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{2 π}{3} + 2 πn,

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{2 π}{3} + 2 πn

into

2 sin (x - \frac{π}{6}) \Rightarrow y = 2

Maximum (\frac{2 π}{3} + 2 πn, 2)

Plug the extreme point

x = \frac{5 π}{3} + 2 πn

into

2 sin (x - \frac{π}{6}) \Rightarrow y = - 2

Minimum (\frac{5 π}{3} + 2 πn, - 2)

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

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

f (x, y) = ln (x^{2} + 4 y^{2} - 1)

x^{3} - 6 x^{2} + 9 x + 5

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

f (x, y) = 6 - x - 2 y

What is the extreme 2sin(x-pi/6) ?
The extreme 2sin(x-pi/6) is Maximum((2pi)/3+2pin,2),Minimum((5pi)/3+2pin,-2)