What is the extreme f(x)=3sqrt(3)x+6cos(x) ?

The extreme f(x)=3sqrt(3)x+6cos(x) is

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extreme f(x)=3sqrt(3)x+6cos(x)

extreme points

f (x) = 33 x + 6 cos (x)

Maximum (\frac{π}{3} + 2 πn, 3 π + 63 πn + 6 cos (\frac{π}{3} + 2 πn)), Minimum (\frac{2 π}{3} + 2 πn, 23 π + 63 πn + 6 cos (\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

33 x + 6 cos (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{3} + 2 πn,

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{π}{3} + 2 πn

into

33 x + 6 cos (x) \Rightarrow y = 3 π + 63 πn + 6 cos (\frac{π}{3} + 2 πn)

Maximum (\frac{π}{3} + 2 πn, 3 π + 63 πn + 6 cos (\frac{π}{3} + 2 πn))

Plug the extreme point

x = \frac{2 π}{3} + 2 πn

into

33 x + 6 cos (x) \Rightarrow y = 23 π + 63 πn + 6 cos (\frac{2 π}{3} + 2 πn)

Minimum (\frac{2 π}{3} + 2 πn, 23 π + 63 πn + 6 cos (\frac{2 π}{3} + 2 πn))

Maximum (\frac{π}{3} + 2 πn, 3 π + 63 πn + 6 cos (\frac{π}{3} + 2 πn)), Minimum (\frac{2 π}{3} + 2 πn, 23 π + 63 πn + 6 cos (\frac{2 π}{3} + 2 πn))

f (x) = f (x, y) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x) = x (20 - 39 + 2 x) (\frac{39}{2} - x)

\frac{x ^{2} + 1}{x ^{2} - 9}

f (x) = \frac{x ^{2} - 6 x + 9}{x - 7}

f (x) = x 21 - x

What is the extreme f(x)=3sqrt(3)x+6cos(x) ?
The extreme f(x)=3sqrt(3)x+6cos(x) is