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

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

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extreme f(x)=2sin(x)+2cos(x)

extreme points

f (x) = 2 sin (x) + 2 cos (x)

Maximum (\frac{π}{4} + 2 πn, 22), Minimum (\frac{5 π}{4} + 2 πn, - 22)

Solution steps

Find the critical points:

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

Domain of

2 sin (x) + 2 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

2 sin (x) + 2 cos (x) \Rightarrow y = 22

Maximum (\frac{π}{4} + 2 πn, 22)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

2 sin (x) + 2 cos (x) \Rightarrow y = - 22

Minimum (\frac{5 π}{4} + 2 πn, - 22)

Maximum (\frac{π}{4} + 2 πn, 22), Minimum (\frac{5 π}{4} + 2 πn, - 22)

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

f (x) = x^{3} - 3 x + 7

f (x) = y^{4} - 4 y^{3} + 2 x^{2} + 8 x y

f (x) = x^{2} e^{- 9 x}

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

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