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

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

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

extreme points

y = x + 2 cos (x), 0 \leq x \leq 2 π

Minimum (0, 2), Maximum (\frac{π}{4}, \frac{π}{4} + 1), Minimum (\frac{3 π}{4}, \frac{3 π}{4} - 1), Maximum (2 π, 2 π + 2)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{4}, x = \frac{3 π}{4}, x = 2 π

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{4},

Decreasing

: \frac{π}{4} < x < \frac{3 π}{4},

Increasing

: \frac{3 π}{4} < x < 2 π

Plug the extreme point

x = 0

into

x + 2 cos (x) \Rightarrow y = 2

Minimum (0, 2)

Plug the extreme point

x = \frac{π}{4}

into

x + 2 cos (x) \Rightarrow y = \frac{π}{4} + 1

Maximum (\frac{π}{4}, \frac{π}{4} + 1)

Plug the extreme point

x = \frac{3 π}{4}

into

x + 2 cos (x) \Rightarrow y = \frac{3 π}{4} - 1

Minimum (\frac{3 π}{4}, \frac{3 π}{4} - 1)

Plug the extreme point

x = 2 π

into

x + 2 cos (x) \Rightarrow y = 2 π + 2

Maximum (2 π, 2 π + 2)

Minimum (0, 2), Maximum (\frac{π}{4}, \frac{π}{4} + 1), Minimum (\frac{3 π}{4}, \frac{3 π}{4} - 1), Maximum (2 π, 2 π + 2)

f (x, y) = y^{3} - x^{3} - 2 x y + 6

f (x) = x^{3} - 3 x^{2} - 9 x + 12

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, (0, 1)

f (x, y) = x^{2} + y^{2} + 12 x - 16 y

f (x, y) = \frac{3}{x ^{2} + y ^{2} - 9}

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