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

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

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

extreme points

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

Minimum (0, 1), Maximum (\frac{π}{12}, \frac{π}{12} + \frac{3}{2}), Minimum (\frac{5 π}{12}, \frac{5 π}{12} - \frac{3}{2}), Maximum (π, π + 1)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{12}, x = \frac{5 π}{12}, x = π

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{12},

Decreasing

: \frac{π}{12} < x < \frac{5 π}{12},

Increasing

: \frac{5 π}{12} < x < π

Plug the extreme point

x = 0

into

x + cos (2 x) \Rightarrow y = 1

Minimum (0, 1)

Plug the extreme point

x = \frac{π}{12}

into

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

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

Plug the extreme point

x = \frac{5 π}{12}

into

x + cos (2 x) \Rightarrow y = \frac{5 π}{12} - \frac{3}{2}

Minimum (\frac{5 π}{12}, \frac{5 π}{12} - \frac{3}{2})

Plug the extreme point

x = π

into

x + cos (2 x) \Rightarrow y = π + 1

Maximum (π, π + 1)

Minimum (0, 1), Maximum (\frac{π}{12}, \frac{π}{12} + \frac{3}{2}), Minimum (\frac{5 π}{12}, \frac{5 π}{12} - \frac{3}{2}), Maximum (π, π + 1)

f (x) = x^{3} - 6 x + 1

3 x^{3} + y^{2} - 9 x + 4 y

f (x) = 5 x^{6} - 6 x^{5} + 1

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

f (x) = 4 x^{4} - 2 (4)^{2} x^{3}

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