What is the extreme f(x)=-2cos(x)-sqrt(3)x ?

The extreme f(x)=-2cos(x)-sqrt(3)x is

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extreme f(x)=-2cos(x)-sqrt(3)x

extreme points

f (x) = - 2 cos (x) - 3 x

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

Solution steps

Find the critical points:

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

Domain of

- 2 cos (x) - 3 x : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{3} + 2 πn,

Increasing

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

Decreasing

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

Plug the extreme point

x = \frac{π}{3} + 2 πn

into

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

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

Plug the extreme point

x = \frac{2 π}{3} + 2 πn

into

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

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

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

\frac{6 x}{x ^{2} + 1}

f (x) = - 1 - x^{2}, 0 \leq x \leq 1

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

x^{2} + \frac{400}{x}

0.05 x + 20 + \frac{125}{x}

What is the extreme f(x)=-2cos(x)-sqrt(3)x ?
The extreme f(x)=-2cos(x)-sqrt(3)x is