What is the extreme f(x)=4sqrt(3)x+8cos(x) ?

The extreme f(x)=4sqrt(3)x+8cos(x) is

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extreme f(x)=4sqrt(3)x+8cos(x)

extreme points

f (x) = 43 x + 8 cos (x)

Maximum (\frac{π}{3} + 2 πn, 43 (\frac{π}{3} + 2 πn) + 8 cos (\frac{π}{3} + 2 πn)), Minimum (\frac{2 π}{3} + 2 πn, 43 (\frac{2 π}{3} + 2 πn) + 8 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

43 x + 8 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

43 x + 8 cos (x) \Rightarrow y = 43 (\frac{π}{3} + 2 πn) + 8 cos (\frac{π}{3} + 2 πn)

Maximum (\frac{π}{3} + 2 πn, 43 (\frac{π}{3} + 2 πn) + 8 cos (\frac{π}{3} + 2 πn))

Plug the extreme point

x = \frac{2 π}{3} + 2 πn

into

43 x + 8 cos (x) \Rightarrow y = 43 (\frac{2 π}{3} + 2 πn) + 8 cos (\frac{2 π}{3} + 2 πn)

Minimum (\frac{2 π}{3} + 2 πn, 43 (\frac{2 π}{3} + 2 πn) + 8 cos (\frac{2 π}{3} + 2 πn))

Maximum (\frac{π}{3} + 2 πn, 43 (\frac{π}{3} + 2 πn) + 8 cos (\frac{π}{3} + 2 πn)), Minimum (\frac{2 π}{3} + 2 πn, 43 (\frac{2 π}{3} + 2 πn) + 8 cos (\frac{2 π}{3} + 2 πn))

f (x, y) = x \cdot e^{- 2 x^{2} - 2 y^{2}}

f (x) = 4 - 7 x^{2}, - 4 \leq x \leq 2

f (x, y) = 4 + x^{3} - 3 x y + y^{3}

f (x) = 375 t - 5 t^{3}

f (x, y) = x^{3} - y^{3} + 3 x y

What is the extreme f(x)=4sqrt(3)x+8cos(x) ?
The extreme f(x)=4sqrt(3)x+8cos(x) is