What is the extreme y=6sqrt(3)x+12cos(x) ?

The extreme y=6sqrt(3)x+12cos(x) is

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extreme y=6sqrt(3)x+12cos(x)

extreme points

y = 63 x + 12 cos (x)

Maximum (\frac{π}{3} + 2 πn, 63 (\frac{π}{3} + 2 πn) + 12 cos (\frac{π}{3} + 2 πn)), Minimum (\frac{2 π}{3} + 2 πn, 63 (\frac{2 π}{3} + 2 πn) + 12 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

63 x + 12 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

63 x + 12 cos (x) \Rightarrow y = 63 (\frac{π}{3} + 2 πn) + 12 cos (\frac{π}{3} + 2 πn)

Maximum (\frac{π}{3} + 2 πn, 63 (\frac{π}{3} + 2 πn) + 12 cos (\frac{π}{3} + 2 πn))

Plug the extreme point

x = \frac{2 π}{3} + 2 πn

into

63 x + 12 cos (x) \Rightarrow y = 63 (\frac{2 π}{3} + 2 πn) + 12 cos (\frac{2 π}{3} + 2 πn)

Minimum (\frac{2 π}{3} + 2 πn, 63 (\frac{2 π}{3} + 2 πn) + 12 cos (\frac{2 π}{3} + 2 πn))

Maximum (\frac{π}{3} + 2 πn, 63 (\frac{π}{3} + 2 πn) + 12 cos (\frac{π}{3} + 2 πn)), Minimum (\frac{2 π}{3} + 2 πn, 63 (\frac{2 π}{3} + 2 πn) + 12 cos (\frac{2 π}{3} + 2 πn))

f (x, y) = x^{2} + 7 y^{2} + x - 6 y + 4

f (x) = 2 x^{2} ln (x) - 19 x^{2}

f (x, y) = x^{2} + 9 x y + y^{2}

f (x) = x^{4} - 12 x^{3} + 4

f (x) = 2 x^{3} + 45 x^{2} - 300 x, 4 \leq x \leq 11