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

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

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

extreme points

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

Saddle (\frac{π}{2} + 2 πn, 0), Minimum (\frac{7 π}{6} + 2 πn, - 3 - \frac{3}{2}), Maximum (\frac{11 π}{6} + 2 πn, 3 + \frac{3}{2})

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: 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{11 π}{6} + 2 πn,

Decreasing

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

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

2 cos (x) - sin (2 x) \Rightarrow y = 0

Saddle (\frac{π}{2} + 2 πn, 0)

Plug the extreme point

x = \frac{7 π}{6} + 2 πn

into

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

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

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

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

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

Saddle (\frac{π}{2} + 2 πn, 0), Minimum (\frac{7 π}{6} + 2 πn, - 3 - \frac{3}{2}), Maximum (\frac{11 π}{6} + 2 πn, 3 + \frac{3}{2})

+ \frac{6 x ^{2} + 35 x - 6}{4 x ^{2} + 23 x - 6}

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

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

\frac{x - 1}{x ^{2} - x ^{3}}

4 x + \frac{8}{x}