What is the extreme f(x)=2sin(x)-cos(2x) ?

The extreme f(x)=2sin(x)-cos(2x) is

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extreme f(x)=2sin(x)-cos(2x)

extreme points

f (x) = 2 sin (x) - cos (2 x)

Maximum (\frac{π}{2} + 2 πn, 3), Minimum (\frac{7 π}{6} + 2 πn, - \frac{3}{2}), Maximum (\frac{3 π}{2} + 2 πn, - 1), Minimum (\frac{11 π}{6} + 2 πn, - \frac{3}{2})

Solution steps

Find the critical points:

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{11 π}{6} + 2 πn

Domain of

2 sin (x) - cos (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Decreasing

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

Increasing

: \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{π}{2} + 2 πn

into

2 sin (x) - cos (2 x) \Rightarrow y = 3

Maximum (\frac{π}{2} + 2 πn, 3)

Plug the extreme point

x = \frac{7 π}{6} + 2 πn

into

2 sin (x) - cos (2 x) \Rightarrow y = - \frac{3}{2}

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

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

2 sin (x) - cos (2 x) \Rightarrow y = - 1

Maximum (\frac{3 π}{2} + 2 πn, - 1)

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

2 sin (x) - cos (2 x) \Rightarrow y = - \frac{3}{2}

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

Maximum (\frac{π}{2} + 2 πn, 3), Minimum (\frac{7 π}{6} + 2 πn, - \frac{3}{2}), Maximum (\frac{3 π}{2} + 2 πn, - 1), Minimum (\frac{11 π}{6} + 2 πn, - \frac{3}{2})

y = x^{3} - 3 x

f (x) = e^{2 x} + e^{- x}

w (x, y) = 3 x + 7 y

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

f (x, y) = x + 2 y