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

The extreme f(x)=2x+cos(2x),0<x<pi is Saddle(pi/4 , pi/2)

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

extreme points

f (x) = 2 x + cos (2 x), 0 < x < π

Saddle (\frac{π}{4}, \frac{π}{2})

Solution steps

Find the critical points:

x = \frac{π}{4}

Domain of

2 x + cos (2 x), 0 < x < π : 0 < x < π

Find intervals:

Increasing

: 0 < x < \frac{π}{4},

Increasing

: \frac{π}{4} < x < π

Plug the extreme point

x = \frac{π}{4}

into

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

Saddle (\frac{π}{4}, \frac{π}{2})

f (x, y) = x^{2} - 3 x y + 5 x - 2 y + 6 y^{2} + 8

f (x) = \frac{9 - x ^{2}}{x}

f (x) = 2 x^{2} - 64 ln (x)

f (x) = e^{x - y} (x^{2} - 2 y^{2})

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

What is the extreme f(x)=2x+cos(2x),0<x<pi ?
The extreme f(x)=2x+cos(2x),0<x<pi is Saddle(pi/4 , pi/2)