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

The extreme f(x)=sin(2x)-cos(2x) is Maximum((3pi)/8+pin,sqrt(2)),Minimum((7pi)/8+pin,-sqrt(2))

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

extreme points

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

Maximum (\frac{3 π}{8} + πn, 2), Minimum (\frac{7 π}{8} + πn, - 2)

Solution steps

Find the critical points:

x = \frac{3 π}{8} + πn, x = \frac{7 π}{8} + πn

Domain of

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

Find intervals:

Increasing

: πn \leq x < \frac{3 π}{8} + πn,

Decreasing

: \frac{3 π}{8} + πn < x < \frac{7 π}{8} + πn,

Increasing

: \frac{7 π}{8} + πn < x < π + πn

Plug the extreme point

x = \frac{3 π}{8} + πn

into

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

Maximum (\frac{3 π}{8} + πn, 2)

Plug the extreme point

x = \frac{7 π}{8} + πn

into

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

Minimum (\frac{7 π}{8} + πn, - 2)

Maximum (\frac{3 π}{8} + πn, 2), Minimum (\frac{7 π}{8} + πn, - 2)

(x - 2)^{2} + 1

f (x, y) = 16 x y - x^{3} - 8 y^{2}

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

f (x) = x^{3} + y^{3} - 9 x y

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

What is the extreme f(x)=sin(2x)-cos(2x) ?
The extreme f(x)=sin(2x)-cos(2x) is Maximum((3pi)/8+pin,sqrt(2)),Minimum((7pi)/8+pin,-sqrt(2))