What is the extreme y=4sqrt(3)x+8cos(x),0<= x<= 2pi ?

The extreme y=4sqrt(3)x+8cos(x),0<= x<= 2pi is

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extreme y=4sqrt(3)x+8cos(x),0<= x<= 2pi

extreme points

y = 43 x + 8 cos (x), 0 \leq x \leq 2 π

Minimum (0, 8), Maximum (\frac{π}{3}, \frac{4 π}{3} + 4), Minimum (\frac{2 π}{3}, \frac{8 π}{3} - 4), Maximum (2 π, 83 π + 8)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{3}, x = \frac{2 π}{3}, x = 2 π

Domain of

43 x + 8 cos (x), 0 \leq x \leq 2 π : 0 \leq x \leq 2 π

Find intervals:

Increasing

: 0 < x < \frac{π}{3},

Decreasing

: \frac{π}{3} < x < \frac{2 π}{3},

Increasing

: \frac{2 π}{3} < x < 2 π

Plug the extreme point

x = 0

into

43 x + 8 cos (x) \Rightarrow y = 8

Minimum (0, 8)

Plug the extreme point

x = \frac{π}{3}

into

43 x + 8 cos (x) \Rightarrow y = \frac{4 π}{3} + 4

Maximum (\frac{π}{3}, \frac{4 π}{3} + 4)

Plug the extreme point

x = \frac{2 π}{3}

into

43 x + 8 cos (x) \Rightarrow y = \frac{8 π}{3} - 4

Minimum (\frac{2 π}{3}, \frac{8 π}{3} - 4)

Plug the extreme point

x = 2 π

into

43 x + 8 cos (x) \Rightarrow y = 83 π + 8

Maximum (2 π, 83 π + 8)

Minimum (0, 8), Maximum (\frac{π}{3}, \frac{4 π}{3} + 4), Minimum (\frac{2 π}{3}, \frac{8 π}{3} - 4), Maximum (2 π, 83 π + 8)

f (x) = x^{3} - 6 x^{2} - 63 x + 7, - 4 \leq x \leq 8

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

f (x, y) = 4 x^{2} + 3 x y + 5 y^{2} + 5 y + 3 y

f (x) = \frac{( 5 x ^{2} )}{ln ( x )}

f (x) = x^{3} - 2 x^{2} + 14

What is the extreme y=4sqrt(3)x+8cos(x),0<= x<= 2pi ?
The extreme y=4sqrt(3)x+8cos(x),0<= x<= 2pi is