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

The extreme f(x)=sin(x)+cos(x),0<x<2pi is Maximum(pi/4 ,sqrt(2)),Minimum((5pi)/4 ,-sqrt(2))

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

extreme points

f (x) = sin (x) + cos (x), 0 < x < 2 π

Maximum (\frac{π}{4}, 2), Minimum (\frac{5 π}{4}, - 2)

Solution steps

Find the critical points:

x = \frac{π}{4}, x = \frac{5 π}{4}

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{4},

Decreasing

: \frac{π}{4} < x < \frac{5 π}{4},

Increasing

: \frac{5 π}{4} < x < 2 π

Plug the extreme point

x = \frac{π}{4}

into

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

Maximum (\frac{π}{4}, 2)

Plug the extreme point

x = \frac{5 π}{4}

into

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

Minimum (\frac{5 π}{4}, - 2)

Maximum (\frac{π}{4}, 2), Minimum (\frac{5 π}{4}, - 2)

f (x) = sin (x), - \frac{π}{2} \leq x \leq \frac{5 π}{6}

f (x) = x + 2 y

y = - x^{2} + 4 x

f (x) = - \frac{3 x}{x ^{2} + 8}

f (x, y) = 50 y^{2} + x^{2} - x^{2} y

What is the extreme f(x)=sin(x)+cos(x),0<x<2pi ?
The extreme f(x)=sin(x)+cos(x),0<x<2pi is Maximum(pi/4 ,sqrt(2)),Minimum((5pi)/4 ,-sqrt(2))