What is the extreme-(sqrt(3))x+sin(2x) ?

The extreme-(sqrt(3))x+sin(2x) is

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extreme-(sqrt(3))x+sin(2x)

extreme points

- (3) x + sin (2 x)

Maximum (\frac{π}{12} + πn, - 3 (\frac{π}{12} + πn) + sin (2 (\frac{π}{12} + πn))), Minimum (\frac{11 π}{12} + πn, - 3 (\frac{11 π}{12} + πn) + sin (2 (\frac{11 π}{12} + πn)))

Solution steps

Find the critical points:

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

Domain of

- 3 x + sin (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{12} + πn,

Decreasing

: \frac{π}{12} + πn < x < \frac{11 π}{12} + πn,

Increasing

: \frac{11 π}{12} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{12} + πn

into

- 3 x + sin (2 x) \Rightarrow y = - 3 (\frac{π}{12} + πn) + sin (2 (\frac{π}{12} + πn))

Maximum (\frac{π}{12} + πn, - 3 (\frac{π}{12} + πn) + sin (2 (\frac{π}{12} + πn)))

Plug the extreme point

x = \frac{11 π}{12} + πn

into

- 3 x + sin (2 x) \Rightarrow y = - 3 (\frac{11 π}{12} + πn) + sin (2 (\frac{11 π}{12} + πn))

Minimum (\frac{11 π}{12} + πn, - 3 (\frac{11 π}{12} + πn) + sin (2 (\frac{11 π}{12} + πn)))

Maximum (\frac{π}{12} + πn, - 3 (\frac{π}{12} + πn) + sin (2 (\frac{π}{12} + πn))), Minimum (\frac{11 π}{12} + πn, - 3 (\frac{11 π}{12} + πn) + sin (2 (\frac{11 π}{12} + πn)))

in v erse f (x) = 1 - lo g_{5} (x - 2)

p er i o d i c i t y f (x) = 2 cos (\frac{2 π x}{3} + \frac{3}{5})

in v erse f (x) = 16 - x^{2}

a sy m pt o t es f (x) = \frac{x + 1}{x ^{2} - 3 x - 4}

in v erse f (x) = lo g_{4} (16 x)