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

The extreme sin(2x)+cos(2x) is Maximum(pi/8+pin,sqrt(2)),Minimum((5pi)/8+pin,-sqrt(2))

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

extreme points

sin (2 x) + cos (2 x)

Maximum (\frac{π}{8} + πn, 2), Minimum (\frac{5 π}{8} + πn, - 2)

Solution steps

Find the critical points:

x = \frac{π}{8} + πn, x = \frac{5 π}{8} + πn

Domain of

sin (2 x) + cos (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{8} + πn,

Decreasing

: \frac{π}{8} + πn < x < \frac{5 π}{8} + πn,

Increasing

: \frac{5 π}{8} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{8} + πn

into

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

Maximum (\frac{π}{8} + πn, 2)

Plug the extreme point

x = \frac{5 π}{8} + πn

into

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

Minimum (\frac{5 π}{8} + πn, - 2)

Maximum (\frac{π}{8} + πn, 2), Minimum (\frac{5 π}{8} + πn, - 2)

f (x) = \frac{x}{( x ^{2} + 4 )}

f (x) = - 0.5 x^{2}

f (x) = 5 x e^{- 5 x}

y = x^{3} - 3 x^{2} - 9 x

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

What is the extreme sin(2x)+cos(2x) ?
The extreme sin(2x)+cos(2x) is Maximum(pi/8+pin,sqrt(2)),Minimum((5pi)/8+pin,-sqrt(2))