What is the extreme y=sin(x)+cos(x) ?

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

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extreme y=sin(x)+cos(x)

extreme points

y = sin (x) + cos (x)

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

Solution steps

Find the critical points:

x = \frac{π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Domain of

sin (x) + cos (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{4} + 2 πn,

Decreasing

: \frac{π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn,

Increasing

: \frac{5 π}{4} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{4} + 2 πn

into

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

Maximum (\frac{π}{4} + 2 πn, 2)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

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

Minimum (\frac{5 π}{4} + 2 πn, - 2)

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

f (x) = 4 x^{2} + 40, - 1 \leq x \leq 1

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

y = x 64 - x^{2}

f (x) = \frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2

f (x) = - 3 x^{2} + 3 x y - 2 x + y^{2} + 6 y + 12

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