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

The extreme f(x)=-x+2cos(x),0<= x<= 2pi is

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extreme f(x)=-x+2cos(x),0<= x<= 2pi

extreme points

f (x) = - x + 2 cos (x), 0 \leq x \leq 2 π

Maximum (0, 2), Minimum (\frac{7 π}{6}, - \frac{7 π}{6} - 3), Maximum (\frac{11 π}{6}, - \frac{11 π}{6} + 3), Minimum (2 π, - 2 π + 2)

Solution steps

Find the critical points:

x = 0, x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 2 π

Domain of

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

Find intervals:

Decreasing

: 0 < x < \frac{7 π}{6},

Increasing

: \frac{7 π}{6} < x < \frac{11 π}{6},

Decreasing

: \frac{11 π}{6} < x < 2 π

Plug the extreme point

x = 0

into

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

Maximum (0, 2)

Plug the extreme point

x = \frac{7 π}{6}

into

- x + 2 cos (x) \Rightarrow y = - \frac{7 π}{6} - 3

Minimum (\frac{7 π}{6}, - \frac{7 π}{6} - 3)

Plug the extreme point

x = \frac{11 π}{6}

into

- x + 2 cos (x) \Rightarrow y = - \frac{11 π}{6} + 3

Maximum (\frac{11 π}{6}, - \frac{11 π}{6} + 3)

Plug the extreme point

x = 2 π

into

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

Minimum (2 π, - 2 π + 2)

Maximum (0, 2), Minimum (\frac{7 π}{6}, - \frac{7 π}{6} - 3), Maximum (\frac{11 π}{6}, - \frac{11 π}{6} + 3), Minimum (2 π, - 2 π + 2)

+ (x - 1)^{2}

f (x) = 2 x^{\frac{5}{3}} - 3 x^{\frac{2}{3}}

y = sin (x) cos (x), 0 \leq x \leq π

f (x, y) = 2 x^{4} - x^{2} + 10 y^{2}

f (x, y) = x^{3} - y^{3} - 6 x y - 4

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