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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme points

f (x) = sin (x) + cos (x), (0, 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, 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)

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